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About Google Book Search Google's mission is to organize the world's information and to make it universally accessible and useful. Google Book Search helps readers discover the world's books while helping authors and publishers reach new audiences. You can search through the full text of this book on the web at |http: //books .google .com/I BIBLIOGRAPHIC RECORD TARGET Graduate Library University of Michigan Preservation Office Storage Number: ACV0278 UL FMT B RT a BL m T/C DT 07/19/88 R/DT 07/19/88 CC STATmmE/Ll 035/1: : | a (RLIN)MIUG86-B35983 035/2: : j a (CaOTULAS)l 60645772 040; : | a MiU | c MiU 100:1 : I a Durell, Fletcher, | d 1859- 245:00: | a Plane and solid geometry, | c by Fletcher Durell. 260: : | a New York, | b Charles E. Merrill co. (c 1911 [cl904] 300/1: : |avi, 7-552p. | b front, (group of ports.) diagrs. |cl9cm. 490/1:0 : | a Durell's Madiematical series 650/1:0: | a Geometry 998: : |cWFA |s9124 Scanned by Imagenes Digitales Nogales, AZ On behalf of Preservation Division The University of Michigan Libraries Date work Began: _ Camera Operator: _ y Google y Google PLANE AND SOLI! GEOMETRY BY FLETCHER DUEEI.L, Pii.D. NEW YORK CHARLES E. MERRILL CO. 44-60 East Twentv-thtrd Street 1911 y Google DuroiPs r^Iathf Plane Geometry 341 pages, 12mo, hiiii Icatlu-r Solid Geometry 213 pages, 12nio, lialf leatliev l^lane and Solid Geometry 614 pages, 12ino, half leather I'lane Trigonometry 184 pages, Svo, oloth . . . Plane Trigonometry and Tallies 298 pages, Svo, uloth . . . P|ane Tiigonometry, with Surveying Tables In preparation Plane and Spherical Trigonometvy, Tables 351 pages, Svo, cloth .... Plane and Spherical Trigonometry, Surveying and Tahles In preparation Logarithmic and Trigonometric Table; 114 pages, Svo, cloth .... Copyright, 101)4, by Cliarles E. Merrill Co. [5] yGoosle PREFACE One of tlie main pui-poses in wi'itiug this book has been to try to present the subject of Geometry so tliat llie pupil shall Tiuderatand it not merely as a series of eorrei^t deductions, but shall realize the value and meaning of its principles as well, This lapect of tlie subject has been directly presented in some places, and it is hoped that it per- vades and shapes the presentation in all places. Again, teachers of Geometry generally agree that the most difficult part of their work lies in developing in pupils the power to work original eseixsises. The second main purpose of the book is to aid in the solution of this difiieulty by arranging original exercises iu groups, each of the earlier groups to be worked by a distinct method. The pupil is to be kept woi-king at each of these groups till he masters the method involved in it. Later, groups of mixed exercises to be worked by various methods are given. In tlie current exercises at the bottom of the page, only such exercises are used as can readily be solved in connection with the daily work. All difflcalt originals are included in the groups of exercises as indicated above. yimilarly, iu the writer's opinion, many of the numeri- m yGoosle cal applications of geometry call for special metliods of solution, and the thorough treatment of such exercises should be taken up separately and systematic ally, [See pp. 304-318, etc.] In the daily extempore work only such numerical problems are included as are needed to make clear and definite the meaning and value of the geometric principles considered. Every attempt has been made to create and cultivate the heiiristic attitiiile on the part of the pupil. This has been done by the method of initiating the pupil into original work described above, by qiieries in the course of proofs, and also at the bottom of different pages, and also by occasional queries in the course of the text where definitions and discussions are presented. In the writer's opinion, the time has not yet come for the purely heuristio study of Geometry in most schools, but it is all-important to use every means to arouse in the pnpil the attitude and energy of original investigation in the study of the siihject. In other respects, the aim has been to depart as little as possible from the methods most generally used at present in teaching Keometi-y. The Practical Applications (Groups 88-91) have been drawn from many sources, but the autliur wishes to ox- prcKs his especial indebtedness to the Committee which has collected the Real Applied Problems published from time to time in Sfhool Science nnd Miitheinatii.-x^ and of which Professor J. F. ilillis of the Francis "W. Parker School of Chicago is the chairman. Page 360 is dne almost en- tirely to Professor AViiHam Eetz of the East High School of Koehester, N. Y. FLETCHER DURELL. LiwaBNCEyiLi.E, N. J., Sept. 1, 1904, y Google TO THE TBACHEK 1. In working original exercises, one of the chief dif- ficulties of pupils lies ill their inability to construct the figure required and to make the particular enunciation from it. Many pupils, who are quite unable to do this preliminary work, after it is done can readils' discover a proof or a solution. In many csereisea in this book the figure is drawn and the particular enunciation made. It is left to the discretion of the teacher to determine for what other exercises it is best to do this for pupils. 2. It is frequently important to give partial aid to the pupil by eliciting the outline of a proof by questions such as the following: "On this figure (or, iu these two tri- angles) what angles are equal, and why?" "What lines are equal, and why?" etc. 3. In many cases it is also helpful to mark iu colored crayon paii's of equal lines, or oE equal angles. Thus, in the figure on p. 37 lines AB and DE may be drawn with red crayon, AO and DF with blue, and the angles A and J> marked by small arcs drawn with green crayon. IE colored craj"ons are not at hand, the homologous equal parts may be denoted by like synibois placed on them, tllUS : F y Google ^ TO THE TE.U'HER (either given or to be proved) by lifi- ures denoting the onk^r in which 11.,/ lines are to be taken. Thus, if 0.1: 0C= 01) OB, the relation iimy !)e indicated as ou the figure. 4. It is sometimes helpful to vaiy the symbolism of the book. Thus, in dealing equalities a convenient symbol for "angle" is 2)^, aa ^ A > 2^1 B. 5. Bach pupil need be required to work only so many originals iu each group as will give him a mastery of the particular method involved. A large number of exercises is given in order that the teacher may have mnuy to select from and may vary the work with successive classes. 6. It is important to insist that the solutions of exer- cises for the first few weeks be carefully written out; later, for many pupils, oral demonstration will be suiilcieiit.and ground can be covered more rapidly "^y its use. 7. In leading pupils to appreciate the meaning of theorems, it is helpful at times to point out that not every theorem has for its oljject the demonstration of a new and unexpected truth (i, e., not all are "synthetic"), but that some theoreips are analytic, it being their purpose to re- duce an obvious truth to the certain few principles with which we start in Geometry. Their function is, therefore, to simplify and clarify the subject rather tbau to extend its content. y Google TABLE OF CONTENTS Definitions akd Fiiwr Piunciplks . . Book I. Rectilinear Fh;uhes .... Boos II. The Cihclb E(.aK III. PKOPORTiaS. SIJKI.AU PoLVt Book IV. Aki:a i-i' PoLvi;(.Na , . , . Book V. Eeri-l.^r Polyiions. Mkahurf KuiiDiutjAL Applications or I'lanl V.\:f Book VI. Likes, Planes and Anolks BofJK Vil. l'OLVHKI)liO\8 Booii VIH, CVE,INDK1:S A.SD C'o.NLiSi . Book IX. The Si'iiiiitE MojiEincAi. Esn;RC(SEs in Somd Geomi MoDEiis (JEOJfETiiic Concepts .... IIlSTOHV OF UHOMl-.THV Review E.'iEiicisi'is in Pi.ani; tlr.oMTyn Review E,\-ehcisi;s in Solid (iEoMEii: I'R.vrncAL Api'i,icatio\s of Ceomf.ti y Google SYMBOLS AND ABBREVIATIONS + plus, or increased hg. Adj. . . adjareiit. — minus, or dimimshed hj. Alt. . alternate. X multiplied hj. Art. . article. -i- divided hy. Ax,. . ax>«m. ~ equals; is (or arc) equal lo. Coiistr. conslrucliox.. A approaches {as a liiiiit] . Cor, . corollary. K= is (or are) equivalent to. Def. . dcfimtion. > is (or ace) greater than. Ex.. . exercise. < is {or are} less than. Est. . exterior. .: there/ere. Fig. . fignrc. X perpejidiCKlar , pcrpeiKUciihir t , Hyp. . hypothesis. or, is perpendicular Id. Ident. identity. Ii.t. . interior. it parallel, or, is parallel lo. I'oBt. . Ila parallels. Prop. . . propositioH. I , i angle, angles m. . . right. A, A triangle, triangles. Sns. . . suggestion. CJ , E£! parallelogran}, paralklogram.i Sup. . supplemciila O, ® circle, circles. St. . . . sfmigbt. q. E. D, quod erat demonsfrandum that is, w hich Wiis 10 proved. Q. E. F. quod erat faeiendum; that i , which was to be ma^ie. A few other abbreriatione and symbols will be introduced i their meaning iodicaied latei: on. y Google DEFINITIONS AND FIRST PRINCIPLES IKTRODOCTORY ILLUSTRATIONS. DEFIHITIOWS 1, Computation of an area, Pr;ieticiil experience lias taught men that cei'fcaiii ways of dealing with objects in the world about us are more advuiifageons tlmn others; thus, if it be desired to find the num- ber of square yards in the area of a floor, we do not mark off the floor iuto actual square yards and count the number of square yards thus made, but pursue the much easier course of measuring two lines, the length and breadth of the floor, obtaiuicg 7 yards, say, as the length, and 5 yards as the width, and multiplying the length by the breadth. The area is thus found to be 35 squai'e yards. Let the pupil determine the area o£ soioe couveuient floor iu oauli of these two waj-s, aud compare the inbof of the two processes. 2. Computation of a volume. Similarly, for example, in order to determine the n amber of cubic feet which a box contains, instead of filling the box with block.s of ■Wood, each of the size of a eubie foot, and counting the number of blocks, we pur- sue the much easier course of measuring the number of linear feet in each inside edge of the box and multiplying together the three dimensions obtained; thus, if 19) yGoosle ]0 PLANE GROHrETE"/ tlie inside dimonMioii,-: an: », ■! anil 3 fci-'t, llif volumu is 5 X 4 X 3, or CO cubic tuat. Similarly, the direet methoii of mensuriiif! the nuniTjer ii£ bushels oE wheat in a bin is to fill a bushel meaaure with wheat from thu bin, time after time, tiH the biu is exhausted, and count the number of times the bushel measure is used. But a much leas laborious method is to measure the three dimensions of the bin in inches and divide their pvoduet by the number of eubic inches in a buabel. Let tlie pupil in like manner compare the lal)or of finding the number of feet of lumber in a given bloek of wood by actually saw- ing the Mock np into lumber feet, with the labor of measni'lng the dimensions of the block a7id computing the number of lumber feet by taking the product of the linear dimensions obtained. 3. Unknown line determined from known lines. Tlie fituiloLit is also probably familiar with the faut tluit, ii.v comjiutiitionB based on the relations of cer- tain lines whose lengths are known, the lengths of other unknown lines may be determined without the labor of measuring these unknown lines. Thus, for instance, if a ladder 5 yards long lean agninst a wall and have ita foot 3 yards from the wall, the hoight of the top of the ladder from the ground may be determined thus: (Height)^ = 5'^ _ 3' = 23 — 9 = IG. 4. Economies in representing surfaces, lines, etc. Other principles of advantage of an even more general eharaeter occur in dealing with geometric objects. Thus, since only one straight line can be passed through two given points, tlie two points may be taken as a highly economized symbol or representative of the line, by the use of which much labor is saved in dealing with lines, and new results are made attainable. y Google riT.nMV.TIilf JtAnNITUDES 11 Similarly, siiieo only one fliit, or plane surf:iee enn be passed through three given points (not iu the same straight line), these three points may be taken as a symbol or representative of the tlat sni-face. This gives the advan- tage not only of reducing an nnlimited surface to three points, but also of giving for the plane a symbol made up of three parts. By varying one of these parts and not the others, tlie plane iriiiy be varied in ouo respeet and not in others; also planes having certain properties in <;onimon may he grouped togetlier, and dealt with iu the groups formed. 5. Geometry as a science. Definition. The above illus- trations serve to Amw limv advantageous it often is to deal with geometric magnitudes by certain methods rather than others. In the study of Geometry as a science we proceed to niake a systematic examination of these methods. Geometry is the science which treats of the properties of continuous magnitudes and of space. GEOMETRIC MAGNITUDES 6. Solids. A physical solid is a portion of matter, as a bioek of wood, an iron weight, or a piece of marble. The portion of space oceiipied by a physieal solid may be considered apart from the physieal solid itself, hence A geometric solid is the portion of space occupied by a physieal solid, or definitely determined in any way. Hence, also, a geometric solid im a limited y)ortiou of space. One advantage id UKing sEOmetrie solids liua in tlda. If we deidt with phyaioal Bolida oiUy, aa blocks oE wood, uuurMu, iron, etc., we should need to determine tlie jiroperties oJ each kind o£ physical aoiid, y Google ]2 riiANE r.KOMF.TKV separately; but, by deterrainlug tiio jiropertits of a geomotric. nolitl, iwe deteimine onee for all the pvoperties of evoiy physical solid, tio matter what its material, that will exactly fill the space occupied by the giveu geometric solid. Hereafter in this book the term "solid" is understood to mean geometric solid, naless it be otherwise specified. 7. Other geometric magnitudes defined as boundaries, A surface is the bonudiiry of a «oiid. A line is the boundary of a surfaiie. A point is the boundary of a line. The solid, surface, line, and point are the fiiijdr.mcntal geometric magnitudes, 8. Geometric magnitudes defined by their dimensions. A solid has three dimensions; viz., length, breadth and A surface has two dimensions, length and breadth. A line has one dimension, length, A point has no dimension. Hence a point is that which has position, but no magnitude. 9. Geometric magnitudes defined as generated by motion. Aline is that which is generated by the motion of a point . A surface is that which is generated by the motion of a line (not moving along itself). A solid is that which is generated by the motion at a surface (not moving along itself). The three independent motions by which a solid is generated ilhia- 10. Geometric magnitudes as intersections. The inter- section of two surfaces is a line; of two lines is a paini. It is sometimes more advantageous to regard geometric magnitudes from one of the above points of view, some- times from another. y Google LINES 13 11. A geometric figure is any combination of poiots, lines, surfaces, or solids. 12. The form or shape o£ .i figure is determined by the relative position of its parts. 13. Similar geometric figures are those whieh have the same shape. Equivalent figures are tliose which have the same size. Equal or congruent figures are those which have the same >jli<tpr and .■<izt\ iind can, therefore, be made to coincide. 14. A point iri represented to tin- eye by a dot nnd ie; niimed by ll lett.et- Liftixed to the dot, as the point A, -A. 15. A straight line is a line such that, if any two points in it be fixed and the line rotated, every point in the line will retain its original position. A straight line is also sometimes described as a line which has the same direction throughout its whole extent; or, as the shortest line connecting two points. The word "line" may be used for — - — ■- "straight line," if no ambiguity results. 16. A curved line is a line no portion of which is straight. The word "curve " is often used for "curved line." 17. A broken line is a line made up of different straight 18. A rectilinear figure is'a figure composed only of straight lines; a curvilinear figure is a figure composed oTily of cnrved lines; a mixtilinear figure is a figure containing both straight and curved lines. y Google .14 PLANE GI,OMKTr 19. Kinds of straight lir.e. A .siraiKlit lin.- dcliTiite Of iudolinitc in Iciij- h. Tho line of (Jefinili^ k'li ;th is Hoiurtiincs b sef/inmi ot- m-cl. Otb^r kiiuta .>f RlraiKht line are .li.fiiicd in Arts. 2C an. 20. Naming a straight line. A straight line i by naming two of its points, as the line AB (a sect) ; or the line CD (in- definite in length). A segment or sect may also be denoted by a single {small) letter, us the Hne a. J" ~ 21. The circumference of a e of whieh id equally dirilaiit fro called the center. -de line every i point w ANGLES 22. An angle i» the amount of opening between two sti'aight linos which meet at a point. The sides of an angle are the lines whose intersection forms tlio iin^jli'; the vertex of an angle is the point in which the sides intersect. 23. Naming angles. (1) The most pr naming an angle is to use three let- ters, one for a point on each side of the angle with the letter at the vertex between these two, as the angle ABC. SiDce the aiae of an angle is indepen- dent of the length of its aides, the points nnmed on its aides may be taken at any plauB on its siilas. Thus, the angles AOl), BOD, iSO£,dOFa.ve all the same auglo. y Google ANGLKfi (2) In fiasc l.liei-c is Init onr iiiigk" at a eivoJi ven the lettor iit the vm-tox alone is sufri«ient to denote angle, as the angie in the last figure. (3) Sometimes a letter or figure phiced inside , the angle and near the vertex is a convenient ziL symbol, as the angle «. 24. A straight angle is an !ii\<;li same straight Hue, hut which ex- tend in opposite direetions fi-oni the vertex, as the ang'le BOA. 25. A right angle is one of two eniiiil one Ktr;tight liiiu hKiuliiig' ;uioll line I'Q meetw lino AB so a.^ to uial'ie angle PQA equal to angle I'QIi, iVM-h of these angles is a right angle. A riy^lit angle is also liiilf of a straight angle. whose sides li.' in the ^ " ef|niil .1 -ill lint nffles made hy . Thus, if the 26. A perpendicular is a line th makes a right angle with a given line; thus PQ in the last figure is perpendicular to liA. The foot of a perpendicular is the point in which the perpen- dicular meets the line to which it is drawn, thus Q is the foot of the perpendicular PQ. 27. An acute angle is an angle less than a right aiiijle, as the angle ^icr. 28. An obtuse angle is a angle greater than a right angl hut less than a straight angle, l angle A OH. y Google IG I'LANE GEOMETP.V 29. A reflex angle is an angle greater than a straight angle, but. less tlian two straigiit angles, as angle ^ Oi^. Ill tliU book, angles larger tliiin a atraigUt angle are not considered unless special mention ia ni;t.i 30. An oblique angle is an angle tliiit is nor a straight iiiigle. Hence, oblique angle is for acute, obtuse, and reflex angles. 31. Adjacent angles are angles whii^li have a common veiiex and a common side between them, an anglt's AOBan<iBOa. 32. Vertical angles are angles which have a common vertex and the sides of one angle the prolonga- tions of the sides of the other angle, as the angles AOG and BOD. 53. Complementary angles are two gotlicr equal a right angle, an the angles AOP a\>d rOQ. Hence, the complement of an angle is the difference between that angle and one right angle. u~ 54. Supplementary angles urn two angles whicii together equal two I'ight /P angles (or a straight angle), as the an- y' gles A OP and POS. / _^ Hence, the supplement of an angle '^ is the difference between that angle and two right angles. y Google ANai.ns 17 85. Angles as formed by a rotating straight line. If the line OB start in the position OA and rotate to the position OjB, it is said to generate tlie LAOB. ? The size of an angle may, therefore, \ I _,-' be considered as the amoniit of rotation ^ _ '^.j, .-'^ ^ of a line about a point, from the origi- /^ nal position of the hue. p/ If the rotation ia eon tinned far enough, a right angle {LAQG) is formed; afterward an obtuse angle {lAOD); then a straight angle {LAOli), and a reflex angle ( lAOV), etc. An advantage of this method of forming or coneeiving angles ia that by continning the rotation of the moving line an angle of indefinite size may be formed. 36. Units of angle. A right angle is a unit of angle useful for many purposes. Sometimes a smaller unit of angle is needed. A degree is one-ninetieth of a right angle; a minute is one-sixtieth of a degree; a second is one-sixtieth of a niinnte. BesidoB tbuse, oDjuv units of angle are used for ceitain puipoaes. SURFACES. DIVISIONS OF GEOMEXEY. PARALLEL IINES 37. A plane is a surface such that, if any two points in the surface be joined by a straight line, the straight line lies wholly in the surface. Hence, it plane figure is a figure sueh that all its points lie in the same j)laiif. 38. A curved surface is a surface no part of which is y Google 18 I'L.VXK f3E0M]':'Jliy 39. Plane Geometry is thnt. bram^li of Geometry wljieli treats of plane figuves. 40. Solid Geometry is that branoli of Geometry wLiiih treats of figures all points of whieh are not in tiie samei plane. 41. Parallel lines are straight lines in tde s;ui;o piitno Tvliicli do not ]neot, liowever fur they be protUiood. Ex. 2 Which of the capital letters o£ the alphabet are str.iight, which curved, which broken, which curved aud straight linta coBihiiied t Kl. 3. Draw an acute angle. An obtuse angle. A refliix angle. Ex. 4. Draw two adjacent angles. ' Two vertical angles. Ex.6. What is the complement of an angle of 43"! What is its Bupplemtnt f Ex. 6, What is the complement of 57° 19' 1 of 62° 23' 43" 1 What is the supplement of each of these ! Ex. 7. At two o'eiocli, what k the angle made by the hour an.l minute hands of a clock 1 at three o'clonk f at five o'.'lo^-k ? Ex. 8. At 1 ;30 o'clock, what anf;lo do the hands of a clock raak,' ? at L>:I5t at 8;-lj? Ex. 9, What kind of an angle is the supplement of . an obtuse angle; of an acute angle f of aright angle t Ex. 10. What kind of a surface is the floor of a roomT the surface of a baseball? the surface of an eggt tlie surface of a hemisphere! Ex.11. If the surface JBCD n the right, what solid is generated h itt What surfaces are generated bounding lines T What tlcos of its angles t ;nerated by ^~~S\ ' ated by its 1 by the vet- Jr=^-:."v.::: ■II ---=.>a y Google GKOMETKIC MAGNITUDES ISI Ex. 12. Draw two supple me ntavy adjacent iicgles. Also two sup- plementary angles that are not adjaeent. Alao two adjacent angles that are not supplementary. Ex. 13. The sum of a right angle and an acute angle iswliat kind of an angle? Their difference is what kind of an angle! Ex. 14. The sum of an obtuse angle and a right angle is what kind of an anglel Their difference is what kind? Ex. 15. If three straight lines meet (but do not intersect) at a point in a plane, how many angles have this point as their common vertes? Draw a figure illustrating this and name the angles, Ex. 16. How many angles are formed if four lines meet ibut do not intersect) at a point in a plane! Es. 17. How many degree ■s art' ii lan auglo whiph equals tw oomj.lement! Ex. 18. How many degret ^s in ai a angle which equals ouf its supplement! Ex. 19. What kind of an angle i s greater than its suppk PRIMARY RELATIONS OF GEOMETRIC MAGNITUDES 42. C<!vtjuii primary relations of geometric objects have already been giveu in the defiiiitioiis uaod for geometric; objecte. We now proceed to investigate the relations of geometrie maenifcudes more generally and systematically. 43. An axiom is a troth accepted as requiring no demonatratioti. 44. Two kinds of axioms are used in geometry: 1. General axioms, or axioms which apply to o.'-.her Ivinds of rinaiitity as wi>ll as to geometric magiiitudesj for instance, to numbers, forcca, masses, etc. y Google liU PLANK GEOIIETHY 2, Geometric axioms, or iisioms \ylii(!h apply to gee- metrii; tiiaguitiidtis alone. 45. The geEcral axioms may be stut.t'd as follows; GENERAL AXIOMS 1. Things H-hieh arc rqiud to fJir siime thintf, or to rqual ihingf, en' rqiiuJ fo Hwli nthrf. 2. Jf rqnah be ciddfid to pcinalu, ihe niim^ are equal. .■J. // equals he subtracted from equals, the remainders are equal. 4. Doubles of equals are equal; or. in general, if equals he multiplied hy equals the pndue/s tire eiii'ul. r>. Halves of equals are equal; or, in general, if eqnah he divided by equals the quotients are equal. 6. The whole is equal to the sum of its parts. 7. The whole is greater than any of its parts. 8. A quaiility may he substituted for its equal in any proeess. 9. Jfeqiials be added to, or suhtr acted from, miequals, the results are unequal in the same order; if uiiequals be added to Uiiequals in the same order, the results are unequal in that order. 10. Doubles, or halves, of uuequals are unequal in the same order, H. If miequals he subtracted from equals, the remainders are unequal in the reverse order. 12. If, of three quantities, the first is greater than the second, and the second is greater tJtan the third, then the jirsi is greater than the third. y Google GEOMETKIO AXIOHR 21 46. The value of the general axioms. Tlie axioms given above seem so obvious that the student at first is not likely to realize their value. This value may be illustrated aa follows: II the distnnee fvom Washington to PhiladelphiB be known, and also the distance fvom Philadelphia to New York, the distance from Washington to New York may be obtained by adding together the two distaneea named; for, by axiom 6, the whole is equal to the sum of its parts. Thus the labor of actually meusuriiig the distauee from ■Washington to New York is savod. Agaiu, if the lu'ifrht ot a suhoolboy o£ a ^'ivon age in Paris be measured, and the height of a like schoolboy in New York be meas- ured, and the result o£ the measnrempiit in each case is the same, we know that tho boys are o£ the same height, without the labor and cost of briugiugtlieboystogether and comparing their heights directly ; for, by axiom 1, things which are equal to the same thing are equal to each other. Thus the general axioms are to be looked at not merely as fundamental equivalences, bitt also as fundamental economies. For many purposes the latter point of view is more important than the former. 47. Tlie geometric axioms may be stated as follows: GEOMETRIC AXIOMS 1. Through Uvo given poinh onhj one slraighf line can be passed. 2. A geometric figure may be freclij moved in space with- out any change inform or she. This aiiom is equivalent to regarding space as mu/orm, or homn- loidal; that is, as having the same properties in all its parts. It has already been assumed in some of the deSnitions given. See Arts. 15 and 3u. 3. Through a given point one straight line a'nd only one can be drawn parallel to another given straight line. By Art. IS, geometric figures which coincide are equal. y Google 22 PLANE GEOMETCr 48. Utility or uses oi the geometric axioms. By means of the firet geonintric axiom we urc able to ishrink or con- dense any striiight line into two points. Later tiiis atlvan- tage gives rise Lo many other advantages. By tlie second geometrie axiom, tbe knowk'djfe whicli ive liaye of ono geo- metric object may be transferred to anotlier like object, how- ever widely separated in space. The utility of the third geometric axiom can be miwle more evident wlieu we come to use it in proving ]iew geometriu truths. 49. A postulate in geometry is a oonstruction of a geo- metric figure admitted as possible. 50. The postulates of geometry may be .-stated as follows : 1. Through any two points a straight line way he dt-airn. 2. A straight Unetnay ie extended indejiHitelij , or it maij be limited at any point. 3. A circumference may be described abont any given point as eenier, and with any given rifdiii.'^. These postulates limit the pupil to tbe hsb of the slrnight-eciKed ruler and tlie compasses in eonatmoting Gbuibb in geometry. One of the objects of the study ot geometry is to disi'over what geometrin figures can be coustructed by a combination of tbe eleiiienfnry eon- atraetions allowed in the postulates ; that is, by the use of tho two simplest drawing instraments. 51. Logical postulates. Besides the postulates which are used in the actual construction of figures, there are certain other postulates which are used only in the processes of reasoning. Thus, for purposes of reasoning, a given angle may be regarded as divided into any convenient num- ber of equal parts. Whether it is possible actually thus to divide this angle on paper by use of the ruler and com- passes, is another question. y Google DEMONSTRATION OF PROPEItTlER '26 DEMOKSTKATION OF GEOMETRIC RELATIONS 52. A geometric proof, or demonstration, is a course of reasoning by whidi a relation lietween geometric objects ia established. 53. A geometric theorem is a statement of a truth con- cerning geometric objects which requires demonstration. Es. The sum of the angles of a triangle equals two ri);'it angles. 54. A geometric problem is a statement of the construc- tion ot a geometric iigiire which is required to be made. Es, On a given line to eonstruet a triangle toutalning three equal 65. A proposition is a general term for either a theorem or a problem. Thus, propositions are sulidiviileil into two classes: 1, Theorenas; 2, Problems. 56. Immediate inference is of two kinds: 1. Changmg the point of view in a given statement. Thus the statement, "two straight lines drawn through two given points must coincide," may be changed to "two straight lines cannot inclose a space." 2. Reasoning which involves but a single step. Ex. "All straight angles are equal;" .'. "All right angles are equal." (Ax. 5.) 57. A corollary is a truth obtained by immediate infer- ence from another truth just stated or proved. 58. A scholium is a remark made upon some particular feature of a proposition, or upon two or more propositiona which are compared. y Google 2-i PLANE GEOMETRY 59. Hypothesis and conclusion. A |iroiiosition consists of two parts : 1. The hypothesis, oi- Unit wlikh is known or granted. 2. The conclusion, or that which is to he ]ii'ovi'd oi- coii- stvueted. Thus, in the pfoposition, "if two straight liiicH are per- pendienlar to the same line, they are parallel," the hypothesis is, that two given lines are perpendienlai" to jinother given line; the eoncUision is, that the two given I'mes are parallel. 60. The converse of a pvoposition is another proposi- tion formed by interchanging the hypothesis and the conclusion of the original proposition. Thus, theorem, "every point in the perpendicular bisector of ii. ]itie 18 equidistant from the estremitiea of the line;" Cotirefse, "every point equidiatant from the extremities of aline lies in the perpendicular biaeetor of the Hue." Or, in geneml, theorem, "if A is B, then A' is F." en»Terse, " if X is F, then A is nr' Frequently a converse is formed by interthanfiing p^i-t only of an hypothesis with part or all of the eoneluuion, or vice vi^rsii. Thus, theorem, "if ^ is B and C ia D, then J is I'r' com-erse, "" if J is B and X is Y, then C ia Jh" The converse of a theorem is not necessarily trne. Tims, it is true that all right angles are equal, but it is not true that all equal angles are right angles. 61. The opposite of a theorem is a theorem formed by making both the hypothesis and the conclusion of the original theorem negative, Ex. theorem, "if A is B, then X is Y;" opposiff, "if A is not B, then X is not 1'." ' y Google DEMONSTRATION OF PROPEETIES 25 If a direct theorem and its opposite ace both true, the uonverge is known to bo true without proof. Also, if a theorem and its converse are both true, the opposite ia known to be true withont proof. Thus certain economlea arise in the demonstration of theorems. 62. Methods of geometric proof. Sevuciil principal methods of proving theorems are used in geometry. 1. Direct demonstration. 2. Proof by superposition, in which two iijjureiS are proved equal by makiiii^ one of Ihein eoiiieide wifh the other. 3. Indirect demonstration, wliieh eonsists essfniially in showing that a given statement is true by i^howiiig that its negative cannot be true. Other special methods of proof will be pointed ont as they occur in the coarse of the work. 63. FornI of a proof. The statement of a theorem and its proof eonsist of certain distinct parts which it is impor- tant to keep clearly in mind. These parts are; 1. The general enunciation, which is the statement of the theorem in general terms. 2. The particular enunciation, or statement of the theorem as applied to a particular figure used to aid the mind in carrying forward the proof. 3. The construction of supplementary parts of the figure (not necessary in all proofs). 4. The proof. This must include a reason for every statement. 5. The conclusion. The letters Q. E. D., standing for "quod erat demon- strandum," and meaning "which was to be proved," are usually annexed at the end of a completed demonstration. y Google PLANE GEOMETi;V EXERCISES. CtiOi Es. 1, InthofiKurempas- ure AB: then meivsure BC. A — _— -a Now find AC without meas- uring it. Wliat axiom hiive you used 1 Ex. 2. If /L ^0^ = 00°, I BOC=m'' and ■'; /''' ZCOi) = 130°; find without measui-ing them a ~X / AOC and BOD (reHe.";). What axiom have you N, / Ex. 3. Prove by repeated use of the first part of Asiom 1 that magnitudes equal to equal [ magnitudes are equal to each other; (thus, I> given J=!.r, li=y, and x~'j. Prove A^B). Ex. 4. Give a namevipal iliuKtration of Axinm 11, Ex. 5. Show liy ttia axioms that a part is equal to a whnlo di- minished by the remaining part, Ex. 6. Show that Axiom 1 is a special chsb of Axiom K, Ex. 7. It a = x + }j and ;c = y, show by use of the axioms that Ex. 8. Di'aw a line and produce it so that the produced part shall equal another given line. Ex. 9. By use of the eompasaes, mark off on a given line a part twieo as long as another given line. Ex. 10. On a givenlinemarkoff.byfoive^t uses of theoompHSses, a part four times as long as another given liue. Ex. 11. Di^aw three straight lines and denote them hy I, ra, andii. Then draw a line l + ni-~n, and also a line I — 2'i(+3ii. PROPERTIES OF LINES INFERRED IMMEDIATELY 64. If two straight lines have two points the lines coitidde throughout their whole extent (Art, 47, Geom, As. 1). Hence, two straight tines can intersect in hut one point. 65. If two straight lines coincide in part, they coincide throughout, y Google PEOPERTIES OP ANGLES 27 66. Only one straight line can be drawn eonnecHng two given points. 67. Two straight lines cannot enclose a surface. 68. A given straight line {sect) can he divided into two equal parts at but one point. For (by Ax. 5) halves of equals (or of the same thing) are equiil. PROPERTIES OF ANGLES INFEEEED IMMEDIATELY 69. All glrnight angles arr equal. 70. A straight angle can he dividfd into lieo equal angles by but one line at a given point . in the given straight Zme. For (Ax. 5) halves of the same mag- nitude are equal. 71. Hence, at a given point in a straight line hut one perpendicular can he erected to the line. 72. All right angles are equal. For all straight angles are equal (Art. 69) and halves of equals are equal (Ax. 5). 73. The sum of the two adjacent angles / formed by one straight line meeting an- F^i other straight line equals ttvo right angles. For the angles formed are supplementary adjacent angles (Arts. 31, 34). 74. If two adjacent angles are together equal to a straight angle {or two right angles), their exterior .Hid«'ts form one and the same straight line. For their esterior sides form a straight angle, and hem;e must lie in a straight line (Art. 24). y Google 28 }'T.ANE 03?,OJIETRY 75. The cowplemeiiix of Iwo equal nn^jlr-: /nr equal (Art, 72 and Ax. 3); the aiipplemetits of two equal angles are equal (Art. Gi), Ax. 3). 76. The sum of all Ike. oiujles about a point equals four rhjhl amjleti. Thus, la + lh + lc + ld^-le--^ 4 rt. A.. ■77. The SUM of all the angles about a point on the same side of a straight line passing through the point equals two right angles. Thus, Zj3 + Z5 + Zr = 2 i-t. i. EXERCISES. CROUP 3 Ei. 1. How many different straight linps are determineii liy tliree points not in the Same straight line? Ex. 2, How many strniifht lines are determined by fouc points in a plane, no three of them being in the same stra:(;(it line ? Ex. 3. If, iii Fig. 2 above, Aa,b,c. cl^ 40°, 50°, 60°, 70° i-esppp- tively, find Z e. Ex. 4. IE,- in Fig. 3 above, the lines forming the an^ld q are per- pendieulat to each other and I p = 47°, find the other angles of Iho fignre. Ex. 5. Measure Zrt o£ Fig. I on preceding page. Find Zb with- out measuring it. Now measure Ih and compare the two results, , A Ex. 6. Given QB1AH, PB 1 BC, and \ ,q IABC = 1W; find tha otliot angles of the \ ^^ figure. \l rrr\' B a Ex. 7. Arrinpe file points in ^ pKne so that the few ! st n imber of st light line'! i av pass through them, no line to pass through more than IhrfiL [ lul? y Google PLANE GEOMETRY Book I RECTILINEAR FIGURES Proposition I. Theoreji 78. If one simight line inUrsecfa anotlipr siraight line, the opposite or vertical angles are equal. GiveE the straight linos AB and CD iutersoetiiig at the point 0. To prove ^AOC= IDOB and IA0T)= ICOli. Proof. /. AOG+ ZA0D=2 rt. A, Art. 73. [tkv Kiim (jf (ICO ndjaecut angles formed hij one striikjht hue meclimj aiiullicr straight line equals tu-o riylit angles). Also ZfiOD+ZJ.OD==2rt. A, isame reason). .-. ZAOC+^AOD=^lBOI>+ZAOD, A^i. i. {things equal lo the same ilibuj are eqiiul to t'udt other). Subtracting ZAOD from the two equals, /.AOG = lBOD, Ax. 3. [ifeijiials he siiblraotcd from equals, the remainders are equal). Id liliu iiiamu;r it may be proved that IA0J) = IC0B. O.B. B. Ex. If i without uihu n the above 6gure IDOB^IQ". isuriug them. Cud the other angles y Google 30 BOOK I. PLAXE GEOMETltV Proposition; II. Tiieoeem 79. If, from a point in a perpendicular' to a given line, two oblique lines be drawn cutting off on the giceu line equ<tl sffimenia from the foot of the perpendicular, the oblique lines are equal and make equal aiujle.s with the perpendicular. Given a line AB witli CI) ± to it at the point C. and FR and FQ drawn from :iny point as P in CD, cutting off CR = QCon AB. To prove FR ~ FQ and l CFR = l CFQ. Proof. Fold over the iigiive DCB abont J}C as an axis till it i;omes in the plane DOA. Geom. Ax. 2. Then ^DCB = /. DCA {allrighl A are = ). Art. T2. .■. line CB will take the dii-eetion of CA. But Hencf And CJi= CQ. .: point R will fall on point Q. ine PR will coincide with line FQ. Hvp. ■u-aiyia ectiu'j U I CFR will coincide with / CFQ. .: FR = FQ, and Z CPE = I CFQ, {geometric figures which coincide are equal). Ex. 1. Point out the hypothesis and the oonelusioii ii eounciatton of Prop, I. Also point thorn out in ti enunciation. Do tho same tor Prop, II. Ex. 2. If three straight lines intersect at a point, 1 the angles formed is it necessary ti all the augleat y Google LINES AND ANGLES 31 pROPOsiTioK III. Theorem 80. From, a given point wUhftut a straight line but one perpendicular can be drawn to the line. Given the str,iif;iit lino AB, P any point witlioiit: AP,, PQ ± .1 /i, iiud PR any othet- line dniwn from /* to .4./-!. To prove that PR is not ± Ali . Proof. ProdneePQ to P' making Q/'' = i*Q. Ih-a.w RP'. Then RQ 1 PP'. Tiyp. P'Q^PQ. Constr. .-. ieP = RJ^ and Z PiiQ = Z P'ii^g, Art. 79. (*/. /'"o»i a ;w/h( t» a ± (0 a ghea line, tvto ohligne lines be drawn cattiiiij o^ on the gii\n line eqiiiil segments from XMftiot of the X,, ike oliliqae lines are eqaal anil make equal A with the L ) . Bnt Pif^F is not a straight line, Art. G6. (only one slniiijht line can be dnuvH eiiiiuecUng tiro gii'en piiiiiln), :. PRI" is not a stmiglit Z . .-. Z PRQ, the half of Z PRP, is not a rijjlit /. . Ax. 10. .-. PR is not ± AP. .'. onlv one perpendicular can be drawn from P to AP, Q. E. ». Ex. Tlll'BB traiRlit liii a iiiterse t tit a pi) aiiglos lormod It tlie iKtiii Me 30° XQd 4U". ftt the point. y Google I'LANE (iKUMF/l'RV TRIANGLES 81. A triangle is a portion of a plane homideil 1 straight liiiow, as the ti-iaiigle AUG. 82. The sidM of a triangle are the lines which bound it; tlie perimeter of a triangle is the sum of the sides; the angles of a tri- angle are the angles formed liy the sides, as the angles ^1, B and G; the vertices of a triaui>:le angles of the triangle. 83. An exterior angle of a triaiigb one side and by another side produced, as the angle BCD. With reference to the BCD, the angles A and B are termed the opposite inte- rior angles. 84. Classiiication of triangles according to relative length of the aides, A scalene triangle is a triangle in whieli no two sides are equal. An isosceles triangle is one in wlueh two sides are equal. An equilateral triangle is one in which alt three sides are equal. y Google TEIANGLES 33 85. Classification of triangles with reference to character of their angles, A right triangle is a triangle one of whose angles is a right angle. An obtuse, triangle is a triangle (me of whose angles is an obtuse angle. An acute triangle is a triangle all of whose angles are acnte angles. An equi- angular triangle is one in which all the angles are equal. 86. The base of a triangle is the side upon which the triangle is supposed to stand, as AB. The angle opposite the base is called the vertex angle, its angle ACB\ the vertex of a, triangle is the vertex of tlie vertex angle of the tri- angle. The altitude of a triangle is die perpend iciil.'ir from the vortex to the ^ base or base extended, as CD. '-' 87. In an isosceles triangle, the legs are the equal siilys, and the base is the remaining side. 88. In a right triangle, the hypotenuse is the side oppo- site the right angle, and the legs are the sides adjaeent to the right angle. 89. Altitudes, bisectors, medians. In any triangle, any side may be taken as the base; hence the altitudes of a triangle are the three perpendiculars drawn cue from each vertex to the side opposite. y Google -1-i BOOK I. I'LANR fiEOMl?T!!Y A bisector of an angle of a triaiiglo is a line wliicli ilividcv tliis angle into two equal pavte. ThJH Liswtur is usiiiLlly pru- diieed to mout tlto side opposite tlio givwi aiiglu. A median of a triangle is a line drawn from a vertex of the triangle to the middle point of the opposite side. How many medians has a triangle ? 90. Two mutually equiangular triangles are triringles having their corresponding angles equal. 91. Homologous angles of two mntnally equiangular triangles are corresponding angles in thnwe triangles. Homologous sides of two mutually equiangular triangJes are sides opposite homologous angles in those triangles. We shall now proceed to determine first, the properties of a single triangle, as far as possible, then those of two triangles. 92. Property of a triangle immediately inferred. The SUM of any two sides of a triangle is giratfir than ihf third mJp. For a straight line is the siiortest line between two points (Art. 15.) Ex. 1, Point out the hvpotiiesis and eonelusion in the generd onun- tiation of Prop, 111 ; also point tLtm out in the pnrtienirLi' enunciation. (As each of the uoxt Cftuen Props, is studivJ, let the pupil do tlia same for it.) Ex. 2. Find tlie angle whose complement is IS"; whose supple- Ei. 3. It the eomplement of an angle is known, what is the shortest way of finding the supplement of the auglef If the supple- ment is known, what is the shortest way of finding the complement ? Ex. 4. In 25 minutes, how many degrees does the minute-hand of a clock travel 1 'Rtm m.iuy docs the honr-hand f Ex. 5. Draw three straight lines ao that they shall i three points; in two points; in one point. y Google TEUNGLES SJ Proposition IV. Theorem 93. Aiiif aide of a triangle is greater than the difference between the other two sides. Given A B any side of the A -4 liC, iiiul A € > iiC. To prove AB>Aa—nC. Proof. AB + BOAC, Art, 93. (the sum ofawj two sitlcs of a triangle ?« i/rcaler lha» the thinl side). Subtracting BG from eacli member of tlie itiequalitj-, AB>A€—BC, A^.!). 'Sf equals he subtracted from unequals, the remainilers are uneijual in the same order). q, s, D. 94. Cor. TJie perpendicular is the shortest line thai can be drawn from a given point to a given line. For, in the Fig. pnge 31, PP'<FB + RI", Art92. Or, 2 1'Q<2rR. Ax. 8. .-. FQ<PR. A>^.V). Hence, Dep. Tlie distance from a point to a line h the perpendicular drawn from tjie point to tlio line. Ex. J, If one sido of an equilateral triniifjle U 4 iiioliea, n-iiat i^ ita perimeter ? Es. 2, la it possible io form a triangle whose sides are 6, £1 and 17 inehea f Tt7 to do this witk the compaaaea and ruler. Ex. 3. Is it possible to form a triangle in whii/ii one side is 10 inches and tho diiferein'o of tLe other two sides is I'l Inches f Ex. 4. On a given line ns base, liy e.xm^t uae of ruler and uoni- pftsaos, eonstrucL au equilateral tviauglo. y Google VLAXE GEOAn'TltV PltDI'OSITlON V. TlIKOliKM 95. If, from a point inth'm a irimu/U, i '■'■■ an ' ''■ nri'm fo the extrfmilies of otie side of the triangle, Ike ^'ihi of ihe other tiro xides of the trimigle is greater than the sum of the two Uii';:^ no drawn. Given P nny FA and I'G lines side AG. lit within tho tvi!\iislfi ABO, aw\ WW from r to the extremities of the To prove AB + BC> AF + PC. Proof. ProaiicR the line AF to meet EC at Q. Tlien AB + BQ> AP+ FQ. A (a siraujIU hue !.s tlie siiortei line coiuieclMg tuo poiul^) I Also FQ + QG> PC, (s,i n>). Adding tliewe inecinnHties, AB + BQ + PQ + QC > AP+ FQ + PC. Substituting BC for its equal BQ + QG, AB+ BC+PQ> AP+ FQ + PG. Bubtraetiiig PQ from each side, AB + BC> AP + PC. Ex. Ozi a given line as base, by exact use of ruler and c onBtvutit au ia03cel«H triangle each of whose lega ia double . E. D. irapassea, y Google TKIANGLES di Proposition VI. Theorem 96. Two triangles are equal if two sides and the in- eluded angle of one are equal, respecthebj, to two sides and the included amjle of the other. id />/;/'' in which AB = I)E, Given tlie triangles ^l HU a AC^BF, and lA^ IP. To prove A ABC = A DEF. Proof. Place the A AB(7iipon the A DEF so that the liue AG iioncide.s with its equal DF. Geoui. Ax. L'. Then the line AB will take the direction of DE, if,.rlA^ZDh!i h,j,>.). Also the point B will fall on E, (for line AE = line DE hj hyp.). Hencfi the line BO will coincide with the line EF, Art. 6G. {i^tlij one stiaiglil line oaH he drawn coimecting two yuiiU:,). :. A ABC and DEF coincide. .-. A ABC = A DEF, Art. 47. (geometric figures lirRicft coincide are equal) . q. E. D. El. 1. What kind of pvoof is used in Prop. VI t (See Art. 02). Ex. 2. IE S A, Baud C=CO°, 70°, 50°, J/J=10, Ji7=19, BC=iS: also Z 7>=00°, I}E=\6, DF^IO: find -i E and F and side EF wit bout measuring tbem. y Google liOOa I. I'l.A VnoroHvnas VII. TiiEOiiKM 97. Ttco ifUmi/l-'s are e-iiml if two anijli'^ 'ind fhr h,. eluded sid<! of one ay cquiil, respet-ticdy, to iim angles and the included side of the other. Given Llic A ABC iiiid DEF in which ^A=ZD, IC= IF, i\uAAG=llF. To prove AAnC=ADEF. Proof. Plaee the A ABG xipon the A I>EF so tliiit AC shall coincide with its equal DF. Gsom. As. 2. Theu AB will take the direction of DF. (for lA^lDhyluji,.), and the point B will fall somewhere on the line 1>E or DF produced. Also the line CB will tal^e the direction of FF, {for LC = lFhjliyp.), and the point B will fall on FE or FF prodnced. .". point B falls on point F, Art. f>4. {two straiglU hues ciin intersect in but one jioiixi]. .'. &. ABG and DEF coincide. .-. AABC^ADEF, Art, 47. {geometric figures toliich coincide are equoX). n e, B Ex. 1. What kind o( proof is used in Prop. VII. ! Ex.2. If A A, B, 0=65°, 55°, GO', AB=2i. Aa=-['i, BC-2T: alio 4 D, f=C5°, 60°, and J)f=18; find DE, EF, and I E, Ex. 3. Conatriict by exiat iisa of ruler and compasses a scalene triaujjle whose sides are 2, 3 and 4 times a givun line. y Google TMANGLES Cif Proposition VIII. Theorem 98. Two right iriangUs are equal if the hijpotenuse and an acute angle of one are equal to the hypotenuse and ait, acute angle of the other. Given the riglit & A BC aiul DEI'^ in wliieii liypote- nuse .4.iJ = liypotoimse DE, iiud lA = Z 1). To prove AABG=£X TiEF. Proof. Pkfie tlie A ABG npon tiie A I>EF so tliat the side AB shall coincide with its equal, the side DE, the point A coinciding with the point D. Then the line AC will take the diret^tion of DF, {for lA^ZDbuhnp.) Also the side BO will coincide with the side EF, Art. RO. {from a given point, E, viithont a straight Itne, DF, but one ± ean be (Irawn to the line). :. A ABC and DBF coincide. :.AABC=A.DEF, Art. 47. {geometric figures ■which cuinckie are eqnal). n equilateral trmngle, n \. E, D. li of whose sidea y Google HOOK I. I'LANE GKOltETRY Proposition IX. Theorem 99. Ill on isosceles trinn'jle the angles opposite- the equal sides are equal. Given the isosceles A ABC in which AB=^BC. To prove ZA = /.C. Proof. Let BD be drawn so as to bisect /.ABC. Then, in the & ABD and DBG, AB^BC. Hyp. Also BD=BI), Ide»t. And IABD= IGBD. Constr. .-. A ABD- A CBB, Art, 96. wo & are eqaiil if two sides and the included Z of one are egual, respee- Uwly, to two sides and the included Z of the other. ] .: /LA=lC, {homologous A of equal A). Ex. 1. On a given line as base, conBtruot exactly an equilateral triangle above the line and another below it. Ex, 2. On a given line as base, eonstruet esactly an isosceles triangle whose leg shall be equal to a given line; make tlie same con- Btrnetion below the given line and join the vertices o£ the two lEOaceles triangles. y Google TEIANCLER 41 Proposition X. Theorem (Converse of Prop. IX) 100. // two ait(]leK of a triangle arc equal, Iha xidei opposite are equal, and the triangle is isosceles. Given Ap frlanfflc ABC in which ^A=-- Z BCA To prove AB = BC. Proof. U thu sides AB and BC are not equal, them must be longer than the otlier. Let AB be thac BC. On ^Sniark off ,10-BO Dmw DO. Then, in the A ABG and ATJC, AD^BC, AC=AG, IBAG^ IBCA. ((wo Si. are equal if iico siijes and the included Z of: reapcctircly, to tico sides and the included L of the otlier) . Ora part is equal to the whole, which is impossible. As. 7, Hence AB cannot be greater than BC. In like manner it may be shown that AB is not less than BC. Hence AB=BC. Art. 90. What ineti.od of lirooE It ill a tri!iiigle VEF, Dtaw u figure nui oil it 1 i usi-d hi Prop. X f ai'fc the value o£ tbe I'ai y Google BOOK I. PLASE GEOHlETIiV Proposition XL Theorem 101. Tim triangles are cqmil if the thrrr .w'.'.s- "f mis •e equal, re^pi'i'lirchj, to the- three sides nf tin- <)(h:r. Given the A ABC iiiul BEF in wliieli AIi = l)E, I1C= J3F, fnidAG^nF. To prove A /U.'C= A BEF. Proof. Place the A ABC so that its longest siile AC shall eoineidu with its equal DF in the A DEF and the vertex B shall fall ou the opposite side of DF from E. Geora. As. 2. Draw the line EB. Then BE^VB (Hy-p.) .: A DEB is isosceles. Def. .-. Ip^ /: r. Art. !)9. (in an isosceles A Wie A opposite the equal sidun are equal). In like manner, in the A BEF, tq=- Ls. Adding, L'P^ Lq= Zr+ Is, Ax. a. Or II)EF=^ Z DBF. Ax. ii. .-. A DEF = A DBF, Art. 9G. {tKo A arc equal if Iwo Hides ami the iHi-hutei Z of one are c<iiial, respcctirclij, to Iwo sides ami tlie inrbuled I of Die other). :. A ABC = A BEF. Ax. 1. — . — Q. E. D, Ex. 1. Construct two equilateral triangles on the same base, one abore and the other helow, and join the two vertices. Prove that the line joining the vertices bisects t1ifi Tf-rie^ iingl"a, and also bisects the base at right angles. Ex. 2. Hence, at any point in a given straight line, construct exactly hy use of ruler and oompasseB a perpendicular to tliat line. y Google TRIANGLES 43 Proposition XII. Theorem 102. Two right triangles are equal if the hypotrniise and a leg of one are equal to the hypotenuse and a leg of the other. Given two right A ABC and Dl^F l!avm<; the hypote- nuse AB =h3-poteimse DE, and BC=J-iF. To prove A AB€= A BEF. Proof. Place the A ABC m that BG shall coincide -with its equal, EF, and A fall on the opposite side of EE from D, at A' . Geom. As, 2. Then A'F anA. ED will form a straight line, A'FI>, Art. 74. (i/ tmo adj. A are toqetticr equal to two rl. A , their ej:t. sides form one and the same straight line). A'E^EIK Hyp. ,■, AA'ED is isosceles. Dt.>f. ;. /1A'= Z.D, Art. 99. ^oscelcs A the A opposite the equal sides are efiiiaL) :. A A'EF= A DEE, Art. 98. are eqiialif the hypotenuse and an amie i. of oi the hypoteniwe and an acute I of the other). AABG=ADEF. a^- __-__^_^~_ '■ ^- "• But (i« ray given stcaigLt Ex. 2. Biaect any yivBii imgk AOB, y Google BOOK I. PLANE GEOMETRY rROl'OKITlOX XIII. ThEOEEM 103. All erierior iniijJe of a iriangk is greater than thcr iippiixite interior anf/k. Given / BGI) an esterior Z. of the A ABC. To prove Z SOD greater than lAJWor /BAG. Proof. Let E be the middle point of the line IIC, Briiw AE and produce it to F, making FE^AE. Draw FC. Then, in tlie A AEB and EEC, AE^FE, and BE - CE, Constr. I BE A= Z. FE€{Mng vertical i). Art. 78. .-. A ^ J:B - A FEC, . Art. %. (frco & are (qual if lico aiih-s and the indiiilril I of one are egtial, re- spet-lk-ciy, to two iiik-s and the indmlcd I of the other). :. ZABE= IFCE, {being bomologmis d of eiiiial A]. But Z BCD is greater than Z FCE, Ax. 7. {the ic/iD?e is greater than any of its parts). Substituting I ABE for its equal ZEOE, Ax. 8. ZBCB is greater than I ABE, that is, than lABG. Similarly, by drawing a line from B through the midpoint of jj (7 :ind by producing SC through (? to a point .ff, it may be shown that lACH {= IBCD) is greater than ABAC. Q- E. D. eacU of wLose legs eqaala balf a given line. y Google TRIANGLES 45 PROPOsiTioN XIV. Theorem 104. If two sides of a triangle are unequal, the angles opposite are unequal, and the grette-" angle is opposite the greater side Given the sida BC > side AB in the A ABO. To prove Z BA C greater than Z ('. Then, in the isosceles AABD, Z)'= Zs, Art. 99, {in an isosceles A the £s opposite the equal sides are equal). ZjB^(7 is greater than Zr, As. 7. [the Kliole is greater than anij of its parts). .■. ZJBAC is greater than Zs. Ax. K. But Z.s is ,111 exterior Z of the A ABC. :. Z/{ is Ki'eater than Z C, Ait. loa. (anext. Z of a A is greater llnm eUltor opposHe iiil. Z ]. Much more, then, is Z BA€ {which is greiiler than Zs) greater than Z C, Ax. 12. (if, iif thrre (fiinniities, the first is greater Ihnn the secmid, OHii the Keeoiiil is greater than the third, then the first is grenler than the Ihu-il) . Q. £. D. 105. Note. The esaeutial afeps of the above proof may be arranged in a single statement, tUua: lJ<AlJ>Zr = ls>lV .: IBAC is greiiter than IC. Ex. J. Which is the longest side o£ a riglit triangle ? of an obtuse triangle f Ek. 2. Conslriiot exactly an equilateral triaugle, Bttth of ivhoee Bides is half a ^iveu line. y Google PLANE CEOME'l'HY PKOi'osiTiON XV. Theorem (Converse op Peop. XIV) 106. If two angles of a triangle are unequal, tie sides opposite are luieqiial, and the greater side is opposite the greater angle. Given I A greiitei- th;iii Z C in the A ABC. To prove BC > AJi. Proof. BC either equals AB, nv is less than .4 B, or greater than AB. Biit BC cannot eqna! AB, for, if it did, /.A would equal Z C, Art,' (being opposite equal sides in an isoscvles A), But this is contrary to the hypothesis. Also BC cannot be leas than AB, for, if it were, Z A would be less than Z <7, Art. i (i/ two sides 0/ n A aye unequal, the i amiosite are uueqattl, ami greats Z ia opposite the tfreater side). This is also contrary to the hypothesis. ;. BOAB, (fzr it ncilhcr equals All, uor in !ras than AH). Q. E. B. El. I. Draw a triangle the altitude of whieh fuUa on the b produced. What kiud of a triangle is tliis i Ex. 2. Draw a triangle the altitude of which coincides with one side. What kiud of a triangle is this f „^ — .^^ Ex. 3. By exact use of the ruler and com- passes, dww a perpendicular to 6 given liue from a given point without the line, ' " y Google TMANGLErt PftorosiTlON XVI. T1IEORH.M 107. If two triangles have two sides of one equal, respec- tively, to two sides of the other, but the included angle, of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. Given tli(! A AIW and J)EF in ivincli AI1 = DE, BC=EF, and lAtW is greater than I.E. To prove AC > liF. Proof, Place the A I)EF so that the side DB coincides with its equal, the side AB, and I*' takes the position F' . Geom, As. 2. Let the line BE bisect the IF'BG and meet the line AC at H. Draw F'H. Then, in the A F-BH and BBC, F'B^BC. Hyp. BH^ BE, Went. IF'BE^ ICBH. ConstT. .-. A F'BH'^ A BEC. (Wby?) .', F'E=CH, {homolot/ous sides 0/ equal ^). Hut AR + EF' > AF", Art. 92. {Die Slim oro'll ''f skies 0/ a A is greater than the third shU], SubslitutiiiK for EF' its equal EC, AE + EC, or AC> AF'. Ax. a. :. AC > 1)F. A.,. 8. q. E. ». Ex. 1. Draw a figure tor Prop, XVI in which the sides and au- gles are of sneh a size that i'" falls within tlie trianftle J fir. Ex. 2. Uraw another figure in whicli *" fiUU on the side AC. y Google 43 llOOK I. PLANE GEOMETRY ritOfORITlON XVII. ThBORE.1[ (CONVERSE OF PUOI'. XVI) 108. // tico sides of a irimigh are equal, respectivehj, to two aides of nnotJier trimigle, but the third side of the. fii-xt is greater than the third side of the second, then the unfile opposite the. third side of the first triangle is greater titan the angle opposite the third side of the second. Given thp A ABC iind HJ^F hiwing AB^-DE. B0= EF, but AC > DF. To prove ZJJ greater tliuii ^E. Proof. The Z.B either equals Z.E, or is less thunZE, or is greater than Z E. But Z B does not equal ZB, for, if it did, A ABC -would = A DEF, Art. or,. {tiro A are eqiial if tico "iilcs and the ineluileil I of one are equid, rcspeclh-etij, to tiro sides ami the iiieluihil Z of the other), and. AC would eqiiiil DF (Jtomoiogoua sides of equal &^) , which is contrary to the hypothesis. Also if AB were less than Z£, side AG would be less than side DF, Art. 107. (if two A liavetwosides of one equal, respeelivelji, io two sides of the olhtr, hut the itictaded Z of the first greater tliait the iacludeil L of the second, then the tAird side of the first is greater than the third side of the second). But this is also contrary to the hypothesis. Hence ZB is greater than ZE, {for it neither equals I E, nor is less than ZE). Q. E. 9. Ei, On a f-iven line (0 con- nictt a triangle whose other two siiles ■e equal to two giyeu lines {m and »). y Google LINES 49 PfiOPERTIES OF tlHES PROVED BY *ISE OF TSIAWGIES PRorosiTioN XVIII. Theorem 109. Of lines drawn from the same point in a perpen- dicular and euttinij off unequal segments from the foot of the perpendicular, the more remote is the greater. Given PO 1 AB, FT and PQ oblique to AB, aud OT > OQ. To prove FT > PQ. Proof. J'roani^e PO to the point P' making OP'=OP. Oil AB take OR=OQ. Draw PE, P'Q. P'R, P'T. Then I'Q = PR, Art. 79. {if^froin a pnint in n ± , a gii-cit line, two oblique linen be drawn, mitliiig off oil the gieai line ri/iiat segmtnts from the foot of the X , the oblique lilies are equal). Ill A PTI", PT+ TP'>PB-VRP', Art. 93. [if, from a point u-illiin n A, two lines be drmnt to the extremilies of a side of the A, the sum of the other two sides of the A is greater than the sum of the tm lines so drawn). But oris X PP and PT Mid P'T cat oft equal segments, PO and i:"0, fi-crni the foot of the L AO. Hence PT^ P'T. In like" manner PR = P'R. Art, 7£». Hence, by substitution, 2 PT > 2 PR. Ax. 8. .-. PT > PR. Ax. 10. .-. PT > PQ. Ax. a. yGoosle PLANE GEOMETRY Proposition XIX. Theorem (Converse of Prop. II) 110. Equal ohlique lines drawn from apoiiU in a per- pendicular cut off equal segments from the fool of ike perpendicular. Given PC ± AB, PK and FT oblique to AB, and PR = FT. To prove CE=Cr. Proof. Ill the right A RFC aiul CPT, PG^PG. Ident. Also PE^PT. Hy,. .-. ARPG^ACI'T, Art. ir- {tico right iJi are equal if ihehijpoleimse ami a leg of one are equal !•• ir,. hypotetiuae and a leg of the other). :. RG=CT, {liovtologotis H'ulei) of eipial ^). 1). E. B. 111. Cor. Of two unequal lines draion from a poini in a perpendicular, the greater line cuts off the greater segment from the foot of the perpendicular. Thus, if PT> PQ (Fig. of Prop. XVIII), OT cannot = OQ (Art. 110) ; nor is OT < OQ (Art. 109) .-. OT > OQ. Hence, also, from a given point only two equal straight lines can be drawn to a given line. y Google LINES 51 Proposition XX. Theorem . 112. I. Efery point in the perpendicular bisector of a line is equally distant from Ihe extremities of the line; and II. Eeery point not in the perpendicular bisector is unequally distant from the extremities of ihe Una, Given the middle point of the line AB, OC ± A P.. P any point in 00, and Q any point not in OC. To prove AP^^PB, bnt QA and QB unequal. Proof. I, AP^PB, Art. 79. (if, from a point in n A- to a gii:en line, tifo ohliqiie lines he drtiwit cutting off OK the given line equal cutji'iinttn, eb:.). II. Since Q is not in the line OG, either AQ or QB must cut the line OG. Let AQ intereeet OG in the point R and join RB. Then AR — RB, [bij first pari of this theorem). To each of these equals add RQ. Then AR + RQ=RB + RQ. Ai. 2. But RB + RQ > qP. .Wiiy?) .-. \>v substitution, AR + RQ, or AQ > QB. Ax. 8. Q. E. D. 113. Cor. Two points each equidistant from the ex- tremities of a line detertnine the perpendicular bisector of ihe line. This corollary gives a useful method of determining tlie perpendicular bisector of a given straight line, by determin- ing two points only of the perpendicular bisector. y Google I'lAKK LiEOMF/niY 114. Dep. Tlio locus of a point is the pnlli of a point moviuL' a<!i;or(lin^ to ii given giiometrii; hiw. Tims, if a point more in a plane so as to be always two iiiehes Jia- tfliit from a given point, its tocua is tlie cirenraEereiiue of a circle wliose center is the given point, and wliose radius is & line two iuPhea in length. TliuB, also, the locus of a point moving so as to bo eqiiidisf!i,nt from two giren parallel liuos is a strait'lit line lyiug midway butween the two given lines. The locus of a point mny p,onKist of two or more sepa- rate lines or parts. Thus, the locus of a point moving so as to be always at a given distance from a f;iven line is two lines, one on either side of tlia given line, at the given distance from it. 115. Demonstration of loci. In order to prove tlint a given line is the locns of a given point moving according to a given geometric law, it is necessary: 1. To prove iJiai every 2>oiiit in the (jireii Hue i^irfisflrs tlie given hnr or condilion. 2. To prox'c that every jwini not in the {/irrn line does not sdiisfif Ihe given law or comHtion. Instead ot 2, it may be proved tiiat every point which sjilisCes t.he given condition lies in the given line. Henee, in Prop, XX it has been proved that the per- pendicular bisector of a line is the locus of all points equi- distant from Die extremities of the line. 116. Use of loci. Loci are useful in determining a point (or points) wLieh shall satisfy two or more geomet- y Google LOCI 53 rieal conditions. For, by finding the loens of all points which satisfy one of^the given conditions, and also finding the locns of all points which satisfy a second condition, and then finding the intersection of these two loci, we obtain the point (or points) which satisfy both coaditious at the same time Thus, if it be required to find the poiuts wliich are two inubes from one glvon poiat and tbrne inchea from another given point, tbe two given points being four inubos apart, the required points are tbe intetsections of tbe circumferences of two t-irtles. Let the pupil inabo a construction and obtu,ia tho riiquiroJ point.H. Ei. 1. Dran inch from a piv f the locUB of rt point movi en point. iiK at the distal ice of one Ek. 2. Drav one inch from a 7 exactly tbo locus of a poii , glvon line. at moving at a distanci 9 of Ex. 3. Driv« distant from the r exactly the locus of a poii 1 extremities of a givou line 1 one iuohlong. ..= .„,. Ex 4 n w tworaralicl 1 n man^ p i ts 1 1 -x plane , OS ' Ihieo panllel hneb ! (■^te irt 4- ) tule 1 Ex 5 \r porrespouJ u„ two t 11 les Ljiil t th Ihr L ingk-* 1 tile tl r e ingl ^ of one t lHufttr.ite by 1. al d iwii the ij, a Ex 6 1 a conne t the i ^ V th ee IS epbIob tiian ks on the same irti es Wb it truth is ilIustratLd 1 y th 3 ijabe figure „d Ex 7 Draw a strait-ht 1 no ind locate a po i t " inches tii By the use of loei locate the po nts wh ch aie 1^^. incbes fi) given line iind aX the same diEtau>,o from the i, ven point y Google 54 BOOK I. I'LANE GEOMETIIY Proposition XXI. Throrkm 117. I. Evpry point in the iiseclor of ini angle diatani from the skies of the angle; aud II. Co^VERBViLV, every point equidixiatit from i of an angle lies in the bisector of the angle. I. Given TB the bisector of the iuieie ABC. P i poii5t hi PB, I'ij J. /:.L, iiiui FB ± no. PQ^FK. lu the rt. A PBQ aad PBB, PB^PB. Id ZPBQ=ZPBE. 1 :. APBQ^APBR, Art, ■c equal if the hypoteimse aii<l ait To prove Proof. Also {tu '- ofo Hence [honiolugotis siJes of equal &). II, Given ZABC, PQ1.AB. PE±BG, and PQ = Pli. To prove tlisit PB is the bisector of /.ABC. Proof. In the right & PBQ and PBS, PB^PB. Ident. Also PQ=PB. Hyp. yGoosle Or .-. A PBQ^-APBE, Art. 102. re equal if the hypolouise and a Icfj of onf, sU:. ) . :. Z AiiP= Z GBP (lio-mhgous A of equal A). ZABCh bisected bv BP. Q. E. D. 118. Cor. In Prop. XSI it has been proved that the bisector of hh ntigk is the hens of all pohils fiquMistanl from the sides of the awjlr, for it h;i5 beon proved that every point in the given Hue siitisfioa tlio given law or condition, and that every point which satisfies thti given condition lies ill the giveLi line (soe Art. 115). E 119. Dr.p. A transversal is a line that intersects two or more other lines. Thus, EF is a trans- versal of the I'nes AB and CD. If two lines are cut by a trans- versal, it is convenient to give the eight angles of intersection special names. n, h, 0, h, are called exterior ;uigles. c, d, e, /, are called interior angles, c, /, form a pair of alternate-interior angles. b, /, form a pair of exterior-interior angles on the same side of the transversal. Let the pupil name another pair of alternate-interior angles; also name another pair of exterior- interior angles on the same side of the transversal; also name a pair of interior angles on the same side of the trans- veraai. y Google 56 iiOOK !. PLANE GKOMETRY PARALLEL LIKES 120. Def. Parallel lines have already iieen ileiined (Art. 4:1) as stmiglit lines wliich lie ic the wimo piano iiiiil do not meet, howcvci' far thoy bu produceil. What is the fuiic!ameutal axiom eonccrniiig pafallel lines? (see Art. 47.) PROrOSITION XXII. TUF-ORKM 121. Two straight lines in Ihesatne plane, perpendicular to the same straight iiite, are parallel. Given the lines AB and CD J. line FQ i\iid in the same plane. To prove AF, || CD. Proof. If AB and CD are oot parallel, tliey will meet if sufficiently prodiieod. Art. 120. We shall then have two Js frura the same point to the line PQ. But this is ii Heace AB and CD never meet. ,■, AB and GD are parallel. y Google PAltALLEL LINES 57 122. Cob. Twosiraight lines parallel to a third straight line are parallel to each other; JAnes parallel to parallel lines are parallel; Lines perpendicular to parallel lines are parallel; Lines perpendicular to noii-paruUel lines are not 2>arallet. Proposition XXIII. Theorkm 123. If a alraii/M line is perpendicular lo mir of iu:o given paraUcl lines it is perpendicular to the other also. Given An II CD, i\w\ PQ L AB. To prove PQ 1 CD. Proof. Let or be drawn X FQ at 0. Then CFWAB. Art. m. IfKo siraigltl lines in the savic plane X snmr. sirnUjhl liiw are ||). But {through n. giccn i<oi CD II AB. CF coincides with CD, I one straight- line, aiul only i aiioOicr gir':a straight iinv). Bnt Hence Hyp. {for CI) coincides with CF, to ii-hich FQ is X)- y Google DO BOOK !. rLANV; Ci;(>?ilETItY Proposition XXIV. Tukokiiji 124. // I'ro parallel sirnigM Uihx irrr ■■nt hij a trans- tersal, llw allcruate inlcrwr <iiighs '.irc equal. Given the 11 lines AB and C7' (;ut by the tt-aiisvevKai I'Q at the points G and E respectively. To prove lAGE^ KjED. Proof. Thi-ongli R, the middle point of EG, let the line ffFbe drawn J. AB. Then HF X 6'/), Art. 1^3. Ufa straight line is X one of tico i] (iiiw, il is X the oiiur alsii). , 111 the right A QJtH and EEF, GE^EB, Consw. /tOMS^-ZEBF. (Whjl) .-. A (Jffiff = A ERF, Art. 98. (too right A (wa ejwa^ ^ the hypotenuse and an aauie Z of one =: (Ae hffpot. and an acute / of Vie other). :. I HGR = Z REF, or, IA0E= I OED, Uiomologous A of equal A). Q. E. B. Es. In tte above figure let the pupi! show that Z BGE = Z GBC. y Google PARAiLEL LINES O'J Prop. XXV. Theorem (Conveese op Prop. XXIV) 125. If two simujhi lines are cut hij a transverml, making the aUenuite interior angles equal, the two straight lines are parallel. ^Given tlie t'wo lines AB and CD cut by tlie triinsveraal PQ at the points F and ff, making Z AFG - I FGI>. To prove AB \\ CI). Proof. Throngh F let the line KL be dmvvn li CI). Then {iftivo II si. Bnt . lines I KFG = Z FGl), are ciU hy a transversal, the all. int. S a IAF6 = IFGD. .\i-t. 124. rv equal). Hyp. Hence IKFQ=- I AFG. :. KL coincides with AB. As. 1. Bnt Heuce (fo> KL II CD. AB 11 CD, Al! •:oincides milt KL, wlurh is \[ CD). Conatr. Ex. 1 . Tf Z BFG = Z FG C, prove that AB aid CD a Ex, 2. By exact use o£ rulai' and eompasaes , lit a given point {P) iu agivenstraigbtline {OA) poivstiiiot an angle equal to a given Z(B), Ex. ; lelti lethods, tbrougti a giv y Google PitOPOSiTiox XXTI. Theoieem 126. 7/ ta-oparnlM lint^s are cut bij a transversal, the exterior interior (Niy/i-s tire eqiuil. Given the || lines AB untl CI) cut by Uie transvprsa! PQ at the points Fand G respectivelj-. To prove ZPFB^ZFOD. Proof. Z FFB = Z AFG. (wiiy! ) IFGD= Z ,-lF(7, Art. 121. Cbeiitg all, uii. A of inirnlld Hiics). .-. IPFB^Z FGI>. As. 1. lu like maniipi- it may be shown that AFFA - Z FGC. q. E. B. Prop. XXVIL Toeori^m (t'oxvEKSK ov Prop. XXVI) 127. If Uvo HtraUjlii linn arc ad J»j n irani:rn-Kal, nuik- ing the exterior interior angles equal, the two siniiqhi lines are parallel. Given, on Fig. of Prop. XXV, ZPFB = IFGD. To prove AB 11 CD. Let the pupil supply the proof. Ex. If, ill l^lie Fig. to Prop. XXVI, /. ['FB equaU €.7°, Ciad tho othi.1 y Google PAEALLEL LINES Gl Proposition XXVIII. Theorem 128. If two parallel lines are cut hy a transversal, the sum of the interior angles oii the same side of the trans- versal is equal to two right angles. Given the ll lines AB and GH ent by the tmnsversal PQ at tlie points F and 6 respectively. To prove Z BFO + IFGI)=2 vi. A . Proof. /iFGI> = ^FFIi. Art. ISO. To VM-], of tliese eciiials add Z BF(i. Then / BFO + IFGD= IPFB + Z BFO. Ak. 2. But Z FFB + Z JiFO^-1 rt. d . A,t, v.i. .-. IBFG + -^FOB^^rt. A. Ax. i. Q. £. D, Pkoi>, XXIX. Thf.(irf.m (CotiVERSF; oe Prop, XXVITl) 129. // iiro straight lines are- nit hy a transversal, maMng the stun of the interior angles on the same side of the trai»iversal hiuuI to tieo right augles, the dm lines are varalkl. Given, on F\<r. of Prop, XXV, Z BFG + Z F0JJ = 2 rt. A . To prove AB \\ CD. Let the pupil supply the proof. Ex. If, ou Fig. of Proi.- XXVIil, ^i'Fli + ^QGD^IBO", are Jli «d CD II r y Google ? BOOK I. PLANE GEOMETKY Proposition XXX. THrcoKicjr 130. Two angles whose sides are paralld, vach to each, •e either equal or supplei. Given A7i || JiR, and EC II GF. To prove A.ABC, DBF and GKIT pqiial, and AABC. and DEG supplementary. Proof. Produce the lines JiC and Hl^ to interseet in P. Then Z ABC = I QPE, and Z QFE = Z DEF, Art. 120. (fceinp eri;i. int. A of || lines). .-. IABG= IDEF. (Wliyf) Also ZDEF= ZGER. CWliy!) .-. lABC^ IGEH. (.WhyO Ajjain .i Ti/JF and BEG are siipplementaiy. Art. 73. .-. A ABC and DEfJ are snppleraentai'v. Ak. 8. Q. E. B. 131. Note. It is to be observed tlmt in tbe above theorem the two angles are equai if, \a the pairs o£ parallel sides, hoth pairs estend in the same direotiou Irom tiie vertices ( i B and DEF), or both paira in opposite directions ( i B aod GEH) i and that they are supplementary it tmepair extends in the same direction, aud the other pair in opposite directions { ii B and DEG). The directions ot the sides are deter- mined by connecting tlie vertices of the angles and observing wliether the liaes considered lie on the same side or on opposite sides of the line drawn. y Google PARALLEL LINES PsoposmoN XXXI. Theorem 132. Two anglfK ; each are either equal a hose sides are perpendicular each to supplementary. Given BA ± EF, and nC X TiD'. To prove ^J.J5C'= IFEI), ami A ABC and FED' sup- plementary. Proof. At E let line EK be drawn ± EF and in tlie same direction with BA ; also EG L !>!>' and in the same direction with BG. Then BA 11 EK, and JJC 11 EG, AH. I2i. (iji'o sirciiijht lilies in tlie same plane, J. (lie same straight line, are H), .-. LABG = /LEEG, Arts, m, ni. ((wo ^ whose sides are ||, eacft to each, aiid extend in the same lUreelii/n from the vertices are =). But I KEG is eoraplement of IBEK, Art. .t^. {for IDEGisart. I lif coiislr.) . Also Z.FET> ia eoniplement of IDEK, Ait. 33. {for I FEE is art. L by conslr.). :. IFED^IKEQ, Art. T5. {,eoT>ipleme»ta of the same L are = ). .-. LABO=LFEJ). Ax. 3. But I FED' is supplement of / FED. Avt, :i4. .-. ZFi'i'' is aupplement of ZAiiC. Ax. 8. Q. E. D, 133. Note. In the above tteorem the two angles ore equal if the sides, oonsidert'd as rot.iting about tiio vertices, are taken in the same order (thus SG ia to the right of BA, and ED to the right of A'."' ■'■Z.ABC= ^FEV); but tlie angles are Bupplementary if theeorre'i- ponding Bides are tiik-flii in tlio opposite onltr (llins,^Cis to the rii:lit ttSAXKHEiy is to tku left of EF :. Z JZ)'C' = Buppleineiit et IFti:!'). y Google PROrOSITIOX XXXII. TuEOTiKJI 134. The Slim, of the. unyles of a Irian.,},' h i'qmil io tiv rUjhi migli's. Given tlifi A A BC. To prove lA+ZIi + l nCA - 2 vt. A . Proof. Pi-odnfie the siile AC totbe point. 7) sniil theon^h C let the line CE be dmwn || AB. Then Z Kr7) + Z iSCB + Z .fiCA = 2 rt. ^ , Art. 77. fi/ic siiiH i)/n!i llie A alHiiit a point on the same side of a su-tiiijlii line pii.'isiiiij tliiojigh the poi>it = 2 rt. S. ). But IBGB^LA, Art. 120. {fe/Hfif (xi. ini. ti. ofpnruUel U»es). Also IBCE = ZR, Art. V2i. {hil,u, nli.int. A ofp'ir„!lfUiw,i). tSiihslitiitiiigfi.r ZICCD itseiiual, /.4,ai)d fnr IBCE it.^ equal, Zli, Ax. 8. /A + zn + ZnVA^'l vt. A. 0- E. D. 135. Cor. 1. An exterior angle of a triungU is eqwil to the sum of the fico opposite interior angles. 136. Cor. 2. The sum of any two angles of a triangle is less than two right angles. 187. CoE. 3. In a right triangle the sum of the two acute angles equals one ri'jht angle. y Google PARALLEL LIN'ES 65 IS8. Cor. i. A triangle can have but one rtgM. or one obtuse angle. 139. Cor. r>. If two angles of one triangle equal two angles of anolher tnangJe, the third angle of the first tri- angle equals the third angle of the second. 140. CoE. 6. If an acute angle of one riyht triangle equals an acute angle of anotlmr right triangle, the rentain- ing acute angles of the triangles are equal. 141. Cor. 7. Tiro triangles are equal if two angles and a side of one are equal to two angles aiul the homologous side of the second. 142. GoR. 8. Two right triangles are equal if a leg and an acute angle of wie are equal to a leg and the Iwmologous acute angle of the other. Ex. 1. If two nngips of tt triiinelG sire M° ami G'i^", find the re- maining angle. Ex.2. If one nciito .nngle of n, riglil trinngle is r>fi° 1"/, find Ihe other acute niiglc, Ex. 3. How xnKny lifgi'Gos in each angle of an equilateval triangle? Ex. 4. IIow nianv degrees in eatli acute angle of an isosceles nsht liianslo! 62", 72°! Ex. 6. If one angle of a triangle is 4^'°, find the sum or the olher two angles. Ex. 7. If two nngles of a triangle are 3S° and 65°. find all the est«rior angles of the triangle. Ex. 8. If tlu^ verti's aiifjie of an isoiicelea trinngls is 3S°, fiml eaeli Ex.9, If an angle at the Use of an i>,llSl^eios triangle is .'iO", find tbe vertes ani;li-, Ex. 10. All f\terior aiisle at tlie liasu of an isosceles triangle is y Google I'LANK <;i:oiiini(Y QUADRILATERALS 143. A quadrilateral iri a portion of a piano boumlod. by four straiglit lines. The sides of a quadrilateral are the Ijounding lines; the angles are the angles made by the bounding lines; the vertices are the vertices of the angles of the quadrilateral. The perimeter of a quadrilateral is the snm oi the sides. 144. A diagonal is a straight line joining two vertices that are not ailjacent. 145. A trapezium is a quadrilaieral no lno o( wlioso sides are parallel. 146. A trapezoid is a quadrilateral whicli lias two, and only two, ot its sides [Kiiallcl. 147. A parallelogram i; sides are parallel. a quadrilateral whose opposite 14S. A rhomboid oblique angles. 149. A rhombus is a rhomboid whose sides are equal. 150. A rectangle is a parallelogram whose angles are right angles. g' 151. A square is a rectangle whose sides are equal. y Google QUADEILATEE.U^S 152. The base of a parallelo- c gram is tlie side upon which it is / supposed to stand, as AB. The / opposite side is called the upper "* base (CD). The altitude of a parallelogram is the perpendicular dia- tance between the buses, as JHF. 153. The bases of a trapezoid are its two parallel sides. The legs of a trapuzoid are the sides which are not parallel. The altitude of a trapezoid is the perpendicular distance he- tweeii the bases. The median o£ a trapezoid is the line join- ing the midpoiuts of the leg:-. 154. An isosceles trapezoid is a trapezoid whose legs Ex. 1. Draw a quiidriltiteval with three acute Ex. 2. Is every rhombus a rhnuitioid J Is thombus ? Bx. 3. \Thatis the ilifferenee b( What [iropeilies do they have Ex. 4. Find the perimeter o£ a square foot Bj Ibid of the following clflBSificatfoii: glee and one ob- ^ery rhomboid a iqwui'e and a vhombuef Quadrilateral ^ Trape^^oid , I. Pavallelogr . Isosceles trapez ) Rhomboid . . . rhombus. Ex. B. Determine what four mimes the rhombus is entitled to. Ex. 6. Datevmiiie what praporties Ihe rbombua, square and rectangle have in commou, Ex. 7. A diagoniil of a rhombus divides tha rhombus into how JUany tviaiiglea F What kiud ot triangles are these { y Google BOOK I. I'LASE (;i:OMETliY Propositiom XSXIII. Theoreji 1 55. The opposite sidi-s of a parallelof/ram are equal, and its opposite an'jtfs (ire also equal. Given the paralklograni ABCI). To prove AJ)=BC, AB^'DC, in=^D, aud lBAJ>-= I BCD. Proof. Draw the diagonal AC. Then, in the A ABC and ADC> AC^AG, (Why?) IBCA = ZCAD, Art. ui. {being alt. int. i of parallel lines). ABAG= AACD, (some reason). :. £^ ABG = C^ AGB, Art. nr. ((H'O A are cqwil if ia-ii A and ihn iiielii/lcd Kide of one an equal rc.iperliri'ly U> liio A aii'l the. incbidvl side of the other). :. AD^BC. A «= BC and Z B= Z 1), ih,moloyo„s pans of equal &). Ill like maimer, hv drawing the diagonal BD, it may be proved that IBAB^IBCD. ^ ^ ^ 156. Cor. 1. A diagonal divides a parallelofjrain into two equal triangles. 157. Cor. 2. ParallelUnes comprehended between par- allel line.i are equal. 158. GoR. 3. Two parallel Uhes are everywhere equi- distant. ^ , Ex. 1. In the above figure, prove ^SJO=ZflCD byuae of Ax. 2. Ejc. 2. Prove tiie oppoeite angles of a, p stall el og ram equal, by use 0£ Art. 130. y Google quadeilateeals; by pROPOSmoN XXXIV. Theorem 159. If the opposite sides of a quadrilateral are equal, thfi figure is a parallelogram. Given the quiuli-il^t.eral ABCT) i F.€=AT>. To prove ABCl) a dl . Proof. Dra-sv the diagonal AG. Then, in the & ABC and ADC, AC=AG. (Why ?) BC^AT). (Why?) AB=CD. (Why 5) .: AABG^AADC. (Why ?) .: Z E.-1C= Z AGT>. (WhyT) :. AB II CJ>, • (ICO lines are ,:i<t h,/ i, Irami-crsal, miihii'i the alt. i IhKl: arc |j). Art. 125. III. i. equal, the Also Z£eA = I CAD. (Why?) :. BG II AD. Art. 12,-K .'. ABOI) is a C^, a ijHUitrilaleral tcliusv ojipo. y Google 70 iXlOli; I. PLANE OEOlIETliY PiiOPOsiTioN XXXV, Theorem 160. // tivo sides of a qiimh-ilateral are equal and par- allel, the other tiro sides are equal andpuraUcl and the figure is a paralletoyrain. Given tlie qiiaaiitaterni A liCD in wliieli BG = and ); AIJ. To prove AB(JD a d/ . Proof. Draw the diagonal .4 C. Then, in U ic A ABC ami ADC, AC=AC. (Why!) BC^AB. (Wby 1) IBCA= ICAT). {bchig •ill. u,l. A of jMuilM Unr^) Art. 124. :. A ABC^AABC. (Why?) .: I BAC=IACB. (Wliy t) tifo Juics arc :. AB\\ CB, Ml by a lnii::<i-fT.i(\l, mildinj Ike all Iwes are [[) . . /(If. A Art. 125. (■.,"ll/, Ilw .-. ABCI) \& 3. rj . ■nt augl. Ai't. 14V J. E. D. Ex, 1. Shon that in a Z35 each pair of adjai:e plementary. Ex. 2. One angle of a parallelogram is 43° ; find the ottier an Ex. 3. If, in the triangle ABC, Z J=60°, Z2i = 70°, whicli i; longBRt Bide in the triangle I Which the ehorteat ! y Google QUADRILATKllAI-S 71 Proposition XXSVI. Theorem 161. The diagonals of a parallelogram bisect each other. Given the diagouuls AC aiul Bl> of the £UAHC'D, intersecting at F. To prove AF=FG, and r.F=FTi. Proof. Let the pupil supply the proof, [SuG. In the A BFC and AFD what si.les uro equal, aiul wby ? What A are equal, and why 1 ett.] £eure V Ex. 2. If one aiiKie III n parallelogram i^ tlu-ee (iiiiL-saQotliev angle, find ali the angles of l!ie parallelogram, Ex. 3. If two angles of a triangle ace j." ami v°, find the third angle. Ex. 4. If two angles of a triangle are j" and OU^ + j°, find the third angle. Ex. 6. If OBB angle of a parallelogram is a°, Ihid thu other uncles. Ex. 6. Construct exactly an angle of W. Ex. 8. liow hirfjB way the doutile of an ubtuab uuyie be ! how y Google l-i 1500K I. I'LAXi; GKOMiyrRV Piim*osrrio.N' XXX VI i. TiiEdiii-.M 162. Two parnJMogramH ui-'' eqind if two ailjticcnt sUhs and tJie included angle of one are, cqii/il, rcspuctinhj, io l/ro adjacent aides and the inchidvd a>i;/l(.' •■fihi' uthn-. Given ihe CD ABV}^i\m\ A'li'Ciy uwyVwh A ti = A' IV, AD = A'I)', and lA =IA'. To prove LTJ ABCD = CI7 A'B'C'D'. Proof. Apply the £17 A'B'G'D' to the CJ ABCT) so that A'D' shall coiueide with its equal AD. Then A'B' will take the direction of AB (for IA'=IA); •and point B' will fall on B (for a'B'=ab). Then B'C and BC will both be 1 1 .-ID and will both pass through the point B. :. B'C will take the direction nf BC, Geora. Ax. .■!. (thro)igh a gircn j>oiiit one siraighi line, ami oiilij one, can he {Irairii \\ aiiolktr gh'cn stmiglit liii':}. In like manner, !>'€' must take the direction of />('. .-. C must fall on C, Avt. C4. (tuto siraighi lines can iittersci-t in hiil one Jioint]. .: Cn ABCI) = CD A'B'diy, Art. 47. (geometric figures it-Mch eoiiicitie are eqval). Q. E. D. 163. Cor. Two rectangles which have equal bases and equal altitudes are equal. Ez. Construct exactly an angle of 30°, y Google 164. A polygon is a portion of a [<lane boutitled by Straight Hues, as ABCDE. The sides of a polygon are its bounding lines; tlio perimeter of a polygon is the sum of its sides; the angles of a polygon are the angles formed by its sides; the vertices of a polygon are the vertices of its angles. A diagonal of a i)o!ygoti is a straight line joining two vertices whkdi are not adjacent, aw IIT> in ¥\^,. 1. 165. An equilateral polygon is a jiolygon all of wliOKe sides are equal, 166. An equiangular polygon is a polygon all of whose angles are equal. What four-sided polygon is equilateral but not eqiii- angular ? Also, what four-sided polygon is both eituilateral and equiaugular ", 167. A convex polygon is a polygon in which no side, if produced, will enter the polygon, as ABODE {Fig. 1). Each angle of a convex polygon is less than two right angles and is called a salient angle. 168. A concave polygon id a polygon in which two or move sides, if prudiK^ed, will enter the pob^gou, as FGUIJK (Kg. 2). y Google /4 HOOK T. PLAXE r.EOilEI-llV Some angle of a ooneave polygon must be greater than two right angles, as angle GlII of Fig. '2. Such an angle is termed a re-entrant angle. If the kind of polygon is not speui&ed in this respect, a convex polygon is meant. 169. Two mutually equiangular polygons are polygons wliorii- eoi't-espoiiding angles are equal, as Figs. 3 and 4. 170. Two mutually equilateral polygons are polygons ■whose corresponding aides ai'e equal, as Figs. 5 and 6. From Figa. 3 and 4 it is seen that two polygons may be mutually equiangular ■without being imituaily equilateral. What similar truth may be inferred from Figs. 5 and 6f 171. Names of particular polygons. Some polygons are used 50 frequently that special names have been given to them. A polygon of three sides is called a triangle ; one of four sides, a quadrilateral ; one of five sides, a pentagon ; of six sides, a hexagon; of seven sides, a heptagon; of eight fiides, an octagon ; of ten sides, a decagon ; of twelve aides, ;i dodecagon ; of fifteen sides, a pentedecagon ; of » sides, an n-gon. Ex, 1, Let the pupil illustrate Arts. 169 and 170 by drawing two pentagons that are mutually equilateral without being mutually equi- aDgalar, and another pair ot which the reverse is true. Ex. 2. Can two triangles he mutually equilateral without being mutually oquianguiar ? What polygons can f Ex. 3. How does the iiumbet of vertices in a polygon uompare witlj the number of aides ! y Google polygons '■> Proposition XSXVIII. Theobrm 172, The sum of the angles of any polygon is equal to iipo right angles taken as many times, less two. as the poly- gon has sides. Given a polj-gon of n sides {the above polygons of 5, 6, 7 sides being used merely as particulai- illustrations, to aid in carryicg forward the proof) . To prove the sura of its A = (ji— 2) 2 rt. A. Proof. By drawing diagonals from one of its vertices the polygon is divided into {n — 2) triangles. Then the sum of the A of each triangle = 2 rt. /^ . Art,i34, (lite mm of the A of a A w eqmt to 3 rl. A ). Heiieethe sum of the Aot the {h— 2)&= (n — 2) 2 rt. .£ . Ax. 4. But the sum of the A of the polygon is equal to the sum of the A of the {ii — 2)^. X-a. S. Ileaee the sum of the A of the polygon = {« — 2) 2 rt. ^ . 173. Cor. 1. Thi> sum of the angles of a polygon equals in— 4) rt. A. eqiiiangidar polygon of n rt. A. yGoosle 7() ]iOOK I. PLANK GEOMl/rnV Pkoi'OSITIOX XXXIX. TuF.oiiKM 175. The sum of llif cxlerim- idi'jU-s of h pohjijoii formeO hif pro'liMinil it.< siik.'i in ^sun-r.-'.'i'ii/ ill cif i:.flri:iiii!ij fijiitii.- four right a>Hjk, Given a polygon of n sides hiivlng iis buIcs pi-oiliiced in suceession. To prove the sum of the exterior A =4 rt, A . Proof. If the iiiterioi" A of the polygon he denotod by A, B, C and the corresponding exterior A by «, h, <\ lA-^ lfi^2 rt. A, (VThj!) l!i + l}) = 2 rt. A , C^'hy ?) etc. Adding, int. A +ext. A = w time.'; 2 rt. :i =2jM-t. ^ . Ax. 2. But int.i = ((/— 2)2rt.:i=2» vi.A — \vi.A. Art. I7:i. .-. Ext. i=4 rt, A. Q. E. D. El. 1. What does the aum of the interior angles of a lieaagon equal ? of a heptagon t of a decagon '. Ex. 2. Each angie of an equiangular pentagon eontaina how many degrees ? of au equiangular hexagon ! octagon ? decagon ! Ex. 3. Would a quadrilateral constructed ot rods hinged tit the ends (i. e., at the verticea of the quadrilateral) be rigid ? Would e, triangle so constructed be rigid J Would a pentagon ( y Google MISCELLANEOUS THE0EEM8 MISCELLAKEOUS THEOREMS Proposition XL. Theorem 176. If three or more parallels intercept equal parts on owe transversal, they intercept equal parts on every trans- versal. Given AP, P,Q, OR nnd T>T parallel lines intercepting equal parts AB, BC and CD on the transversa! AB. To prove that thev intercept equal parts PQ, QR and liT on the transversal FT. Proof. Throngh A, B and C let AF, BO and CR be drawn parallel to PT and meeting the hues BQ, CR and DT in the points F, 6 and H respectively. Then {twos. In the / CDI{, Also Aut But {parallel ii the lines AF, BO and CE are li , Art. 122. aislU hues II a third straight line are || eai:h vther). %. ABF, BCG and CDM, Z ABF = Z BCG = {being ci-i. int. A ti/|| Unm). IBAF^ ICBQ= IDCH, [Mmcreason). AB=BC'=CD. Hyp. .-. AABF=A BCG^ACBH. Art. 97. .-. AF=BG = GR. (Whyn AF^I'Q. liG=QR, CU = RT, Art. 157. ifs 0("i;ii-f/(n(i;«[ bc(ii'p«i. pnraW lines arc equal). .-, PQ^QR^RT. Ax. I. yGoosle PnOPOSlTIOX XIJ. TllEdRElt 177. The line xvUdi join^ flw mhlpnhiU of livo shUs of a triangle is parallel to the third nidu, and is equal io one- half the third side. Given J) the midpoint of AB, and J-J the midpoint < AC in the triangle ABC. To prove BE \\ BG and =iBC. Proof. Through B let BL bo drawn 11 AC, and meetir DE produced at L. Then, in the A ADE and BDL, AD^BD, (Why!) lABE^ZBBL. (Wliyr) ZDAE^ZDBL. (Why?) .: AABE^ABOL. (Why?) .: I'K^JIL, or DE^h BE, and A l-]---BL. (Why?) Bnt, A E ^EG {llyi^.) :. FA'.^-HL. Ax. 1. Al«o EC 11 BL. Constr. :. BLEC is ;i CJ , Art. 160. ' Ucu Hides of a quiulrUuleral arc equal ami ptirtdlH the figio ■eisfl^Z?). :. BE 11 BC. Art. 147. Also LE^BC, {opp. sides ofaCy are =). Art. 155. .■.iLE,OTBE = iBG. As. 5. q. E. I). 178. Cor. The line which hiseds one side of a triangle and is parallel to another side bisects the third side. Thus, given AD = DB and DE 11 BG, then will AE=E<J. For, suppose a line FG drawn through A \\ BG, y Google MISCELLA.MEOUS THEOIUIMS 79 Then the three parallels FG, DE, JiC will intercept equal pai'ts on AB. Hyp. ,". they intercept equal parts on AC. Art. 1T6. .-. AE^EG. Proposition XLIT. Tiieckem 179. The line wkieh joins the midpoints of the legs of a trapezoid is parallel to the bases ami equal to one-half their smn. Given the trapeKoiil ABCB, E the mklpoint, of the leg AB, and F t\\e midpoint of the leg' CB. To prove that a line joining E .iiul /'Is 1| AT) and BC and = i UD+ BC). Proof. Draw the diagonal BI) and take (r its mid- point. Draw EG and GF. Then, in" the A AfW, BG\\ABa\i(\ = A AT). Art. J77. {tU line which Joino f!ic miilpnfiits nf dm ^iili-^ iif a i la [l Ihe (liwd stile and = oiic-half tin: third eiile). Also, iutlie Ai'i^C, GFllliCand^i BC, {sawc r.r^mn). .: OF ami AD hoihwnC; .-.CFwAI), Art. 123. {two striiii/Jii lines \\ a t/iirU stnUnht are \l each oi/iei-): .-. EG and GF are both 1| AD. • '. EG and GF form one and the some straight line EF, ( Oirmigk a giccn point one line, and onhj one, caa be drawi gixcnImcS. :. EF\\ABm(kBC. Also EG = I AB, and GF=l BC. Adding, EG + GF, or EF=^ {AB^BC). 180. Cor. A line draicn hi.'fecfh: and parallel lo fhe base hisccln the oh OilclPf, of, r hij also. mioliier y Google ]iOOt; I. rj-ANE GEOJIKIUY Propositios XLIII. TuEOiiEir Tlu- Inscrtors of thr am/kn 'fa trlimt/ir Given Uie A ABC with tlic lines .1/', BQ, CJi (Fig- 1) bisecting the/. BAC, ABC, ACB, respectively. To prove tlmt AP, BQ, CR intersect in a common point. Proof. Let AP, the bisector of / BAC, aiul CR, tlie bisector of / BCA, intersect at the point 0. Then 0, being in bisector /IP, is equidistant from AB and AC, Art. 117. (every point in the hmctor of nnLU equidistant from Hie tides of IheZ). Also 0, being in bisector CR, is equidistimt from .If" nncT Ii( ', {mini! iT'fXoii). llenuc O h cqiiiOistaiit I'roni tlie riidus ,-l/>' anJ li< '. Ax. 1, . -. (I is in the lilsector of /. A BC, Art. i it. {every point e^idiliMiintfrom the sides of'luZ lies in the hiMVtor of Qia Z.) . Hence BQ, or BQ prodtiueil, passes through 0. Hence the biseetors, AP, BQ, CR, of the three A of the A ABC intersect at 0. Q, E. D. 182. OoR. The point in which the three bisectors of the angles of a triangle intersect is equidistant from the three sides of ike triangle. 1 83. Dbf, Concurrent lines are lines which pass through the same jioint. _„_____ Ex. Find other Z of above flgurs if Z /M f = 73= aud Z BCA =44'. y Google MlSCELLANEOrS THEOREMS Proposition XLIV. T 184. The perpe-ndicular bisectors of the sides of a trian- gle intersect at a common point {called the circumcenter). Given tlie A ABi- with DP, EQ, Fli. (Fig, 1i, Uio j_ biseclufa of tliu sides AB, H<.\ C.\, reKpectivoIy. To prove tiiat T)P, EQ, FR intersect at a comimin point. Proof. Let DP and EQ intersect at the point 0. Then 0, being in _L OP, is eqnidistiiut from tlie jjoiots A and B, Art, 113. {ever-ff poiiU iii, We perjieiulkiilttr bisector of a line in equaily dieivnifrmn ilte exbremiiiee of ike Urie) . Also 0, being in J_ EQ, is eqnidistunt frorii the points B and C, (mme reawu). Hence O is cquidist[uit fram .1 imd ( '. Ax. 1. .-. is in the ± bisector of AC, Ait. 115. {the, \_ hiaector of u line U Hie locus of alt points equidisUfiil from the exireifiitiei! of the line). Henee FR, or FR prodnoed, passes throngli (). Hence the perpendicuhir bisectors, }iP, EQ. FR. <i\' ibc lliree sides of the A ABC meet in tlio point 0. a. e. a. 185. CoJ!. '!'}i<- pi)i))l ill which lin; pi'r}n'iidivuhir hiKi'vlKn^ of Ih,; .u(l,s of a iriuiujl, mcvl is equidisloiit from thi- verliim of the triaiujU: yGoosle [■LANK GEOMF-TRY rilOroSITION' XLV. THEORE^il 186. Tlif perpi'mVculars from the ret-firca of a triangle to the oppogi/e sldca iiuet ut a point {called Ihe ortho-center). 'A N M /^ w Given AD, BF, am\ CB the perpendiculars from the ver- tices A, B, and C of the A ABC to the opposite sides. To prove Ihiit AD, BF, and CE iotorsect in ;i comiiiou point. Proof. Through A, B, and C let the lines PR, VQ and QR be drawn || BC, AC, and AB, respective! y, and forming the A FQR. Then AD 1- VB, Art. 123. (/,»■ AD 1 BC. njjrf r, U:ir J. n:ii; of lli-o [[ !i»rs U 1 the Oilier ulm) . Also A ]' BO imA A BCli are Z17 . Constr. .-. AT = BC, and AB = EC. Art. 155. (((((! opponiic si(}es of a CD are = ). .-. AP = AR. As. 1. lience, in the A FQR, AD is the perpeudicukr bisector of side PR. In like manner it may be shown that BF is the perpen- dicular bisector of FQ, and that GE is the perpendicular bisector of QR. Hence AD, BF, and CB are the perpendicular bisectors of the sides of the A PQB. .: AD, BF, and CE meet in a conimon point, Art. 184, ithi }>ci-pendieula,r biseeiois of the sides of a A arc concurrent). Q- E. B. y Google miscellaneous theokems 83 Proposition XLVI. Theorem 187. The medians of a triangle intersect {or are con' current) in a point {called i?ie centroid) which cuts off two- thirds of each median frotn its vertex. Given AD, BF and CB the three medians of A ABC. To prove that AD, BF and CB intersect at a point which cuts off two-thirds of each median from its vertex. Proof. Let the medians AD and CE intersect at 0. Take R the midpoint of AO, and S the midpoint of OC, and draw RE, ED, DS, and SR. Then ED 11 AC and = I AC. Art. 17T. Also, in the A AOG, lis II AG and - i AC. Art, 177. ED II RS and = ie^'. Art. 122 aad Ax. 1. .-. REDS is a £3' . ES and RD bisect each other. AR^BO, and CS=^SO. .: AR = RO=OD, and CS^^SO^OE. Ak. i. Hence t'£erosses.'IJ.'at a point 0, such that A0=|.4i). In like manner it may be shown that BF crosses AD at the point 0. Uuuite the medians A D, BF, iuid OF intersect at point 0, which cut-s otf two -thirds of each median from its vertex. Q. E. D. Hence Hence ■Bnt (Why!) (Why!) yGoosle 84 BOOK I. TLAKE Cr.OHETEY 188. Properties of rectilinear figures to be proved by the pupil. Proof of equality of triangles. Otlier properties of reetUioear flgares will now be given vvliioh are to be demon- strated by tlie pupil. These theorems wili be arranged in groups, according to the method of proof to be used, followed by a group of general ur mixed exercises. Le.t the pupil form a list of the, conditions that make two triangles equal. (See Arts. 96, 97, 1)8, 101, 102, 141, 142.) £XERCIS£S. CROUP 4 EQUALITY OV 'rilIA.\'Gl.ES Ex, 1. Riven AHC any triangle, BO Iha hi- £ seotor o( I ABC, and JD 1 BO; proved J B0== A BOD. [SUG. In the A AffO and nno wbal Hues are equal! What A are equal t etc.] Ex. 2. If, at any point in the biseolov of an^ie, a X be ereeted and produced to meet t Bides of the angle, how many triangles bvo forint Are these triangles equal ? Prove this. Ex. 3. If, through the midpoint of a f straight line, another line be drawn, and prnduc to meet the perpendiculars erected at the o( of the given line, the triangles so formed i P^ sect each other and their of equal triangles are Ex. 4. H two straight 1. ties be joined, how many Prove this. Ei. 5. It equal segments from the base be of an isosceles triangle, and lines be drawn from segments to the opposite vertices, prove that t angles are formed. y Google ESEECISES. EQUALITY OP TRIANGLES 85 Ex, 6. If, upon the sides of an angle, equal segmnnts be laid off from the vortex, and lines be drawn fram the ends of these Begments to any point in the bisector of the angle, prove that the trianglea formed are equal. Ex. 7. If two sides of a triangle be produced, each its own length, through the vertex in whioh they meet, and the extremitiea of the produced parts be joined, prove that a new triangle is formed whieli equals the original triangle. Ex. 8. Given AB^DC, and BC^DA; yro- ABAC=A1>AC. W triangles is tbare in the figure , and BC^ DA : prove /\g-^ Ex. 9. Two right triangles are equal if ilieir corresponding legs Ex. 10. The ttltitudsH from the extremities of the base of an isosceles triangle upon the legs of tlie triangle divide the tiguve into how many pairs of equal triangles ! Prove this. Ex. 11. In a given quadrilateral two adjacent aides are equal and a diagonal biseots the aogln between tlie^e .■sides. Prove tliat the diagounl bisects the qnadrilaternl. Ex. 12. If, from the ends of the shorter base of an is os eel es trape- zoid, lines he drawn parallel to the legs and produced to meet the other base, prove that a pair of equal triangles is formed. 189. Prooi of the equality of lines. Thei-c are several methods of proving tiint two lines (segments) are equa!. One of the principal tnetJiods of proving that two lines are equal is J>y provitifj that two triangles, in tvMch the given lines form homoloijoit» /uirls, are equal. y Google EXERCISES. CEtOUP a EQUATiiTY OK j.ixr:.-; Ex. 1. Given ABC niiy t.-ismsle, JIO the h)- BPPtorof lABC, and -J/) ± HO; prove AO = On. Ex, 2. K, at any point in tbe bisector of so angle, a porpendicular be ereeted to the bisector and produced to meet the sides of tho angle, the perpendicular 13 divided into two equal parts at Ex. 3. If two aides of a triangle be produced, each its own length, tbronyh the common vertex, the line joining tbe extremities of the produced parts equals tbe third side of tbe triangle (i e., !»£ = BA). Ex. 4. It eqnal segments from the base bi- liijil otT on llie legs of an iaoectles triangle, lined dr;i«n these segments to the opposite vertices are isqual. Ex. 5. Given AR\\ QB, and AF = FJ1: prove BF = PQ. Ex. 6. The bisector of the vertical an- gleof an isosceles triangle bisects the base. '^ ■" Ex. 7. Tiie aititiides of an isosceles triangle upon the eqnr rectangle are equal, isosceles triangle to tho legs nt a triangle are equal, the 1 Ex. 8. The diagonals of Ex. 9. The medians of i Ex. 10. If two altitude- ioaeeles, Ejt. 11. Tbe perpendiculars to a diagonal f a parallelogram from a pair of opposite ertiees are equal. Ex. 12. If the equal sides of an isosceles triangle be prod hrough the vertex so that the produced parts are equal, the lines □g the extremities of the produced parts U> the extremities o lase are equal. Ex. 13. If tbe base of au isosceles triangle be trisected, IrawL from tho vertex to tbe points uf ti'iaectiun are equal. y Google EXEECISES. EQUALITY OF ANGLES 87 Lines may also be proved equal by showing that they are: opposite equal angles in a triangle; or opposite sides of a parallelogram; at parallel lines comprehended hetweeit parallel lines, etc. (See Arts. 100, 155, 157, ete.). Ei. 14. If the exterior angles at the base of a triangle are equal, the trian- Ex. 15. In the biseotora of the equal angles of an isuseftles triangle, the segments nest to the base are equal {AO= OP). Ex 16. In A ABC, given AJ) = AC. DE II BC; prove Ah = AE. Ex. 17. Given AB = DC, a.)iABC = AD; provf, AE= EC. 190. Proof of the equality of angles may be obtained in several different ways. One of the principal methods of proving that two angles are equal is by proving that two triangles, in tvMch the gieen angles form Jyyimlogous parts, are equal. EXERCIGES. CROUP' & EQUALITY OF ANGLES i1. 1- Given J^r: any A, BO the bisector of the lABC, amlJDX/iO; prove lliAO = ^ BDO. Es. 2. If, at any point in the bisector of an angle, a perpendieular bu erected and product;J to me^t the sidea of the anjtle, the ptr- pendieular makes equal angles with the sidett of the angle. Ex.3, Given AH-liV, M^ AD - BV; pruvb IM~/ n. y Google 88 IJOOK I. I'LANE GEOJILTJIV Ex. 4. If equal segments from the base be laid olT on Iho li%'3 of an isoswlea ti'iaiigle, the lines <lrawn from the extreniilips of ibe segments to the opposite vevtii;es miike equal anglus vvilh llie base. Ex. 6. The altitudes upo equal angles with tbts base. Ex. 7, Theai.tgoiia Aiij^los may iilso he proved cqivA aft- opjMsite equal sides in an i^oKcrli-i^ li-ianyh; or uir veriical angles; or ore complements [or supfiUmenlx) of equal ani/les; or hij the use. of the propeHies of parallel lines; or that their sides are parallel, or perpendicular. (Sco Arts. 78, 99, 124, 1120, 130, 132.; I isosBfiles ti-ia,ngle the txteiior auf^li ■ing the base are equal. E*. 9. Given AC = CB, an.t DE Jill; prove Z.CDE = ZCJSn. ',x. 11. Conversely given = Z B, and CE \\ AB ; ■e that CE bisects I DCA Ex. 12. Given BD tbe bisector of thi angle ABC, and PR \\ CB; prove PJIB isoseelea A. Let the pupil state ibis theuri in general language. y Google EXEKCISES. PAEALLEL LINEH hig ihe a'ier- Ex. 13. Given AB = AC, aii3BD=CE; prove ZBCfl= Z CBE. How many pairs of equal A in the figure f of equal lines? of equal A1 Ex. 14. A line drawn through the vertex of an angle, perpendicular to the bisector of the angle, makes equal angles wilh the sides of the given angle. Ex. 15. If a Btraigbt line which biaecis one of two vertical angles be pi'oduued, it biseots the other vei'tioal angle also. 191. Proof that two lines are parallel i bj' showing that: itie Ihu'fi lire cul ha a l7-(ntsi'ernfil, » nate interior angles equal; or making the extenor interior angles equal; or making the interior angles on the same sitle of the transversal supple^iieniary ; 03- that the lines are opposite sides of a parallelogram; or tJuit one of the lines joins the midpoints of tico sides of a triangle, and the other line is the third side of the triangle. (See Arts. 125, 127, 129, 147, 177, ett.} EXERCISES. CROUP 7 PARALLEL LINES Ex. 1. If two sideg of a triangle be produci'ii, each vertex, the line joining their ex- tremities is parallel to the Ihird side of the triangle. Ex, 2. The bisectors of two alternate interior augles of pui- allel lines are parallel. A = ZS, and IDCE = lECA ; -Ex. 3. Given rove CE \\ BA. y Google \n\ HOOK 1. PLANK iu:o.Mi/n;v Ek. 4. The hi life tors o£ the opiiositci aiiflea oC a ]Kir;ilkloi;ram are Ex. 5. l.im.a perpcu.li^iilar lo parallel Imp-. Krs pamllel (or B paiiillfl if Iwo points on oua iiue 192. Tlie proof of a numerical property of the recti- linear figures (of Book I) usually depends on one of the following; The sum of the angles abmit a given point on the sami- side of a straighl line passing through the point is 180 ; or the smn of the angles about a point is 3G0°; <ii' the sum of the angles of a triangle is 180° ; or the .iiiiu of the interior angles of parallel lines on the same side of a transversal is 130°; or file sum of the interior angles of a polygon of n side^: is (ft--2) 180°; :;i- the Slim of the erlertor angles of a polygon is 360°. (See Arts. 76, 77, 128, 134, 172, 175.) EXERCISES. CROUP 8 ^;L■.■^[EKK'AI. rROl'KKTlKS Ex, 1. It an e-ttfirior angle o£ a triaiitcle is "123° and an opp^ inferior angle is SK", find the other two angles of the triangle. Ea. 2. Find the angle formed by the bis'eotoi-B of the two s bugles ot a right triangle, Ex.3. If two angles of a triangle are 50" and 60", find the s formed by their biaaotora. Find the same if the two angles toi Ex.4. It -he vertm aiigle of an irascejes triangle is 4(1" a ;pei'pendieiilar I. drawn from an extremity o£ the base to the opp ■liile, fliiU Ibe ai.^^lea of the figure. y Google EXERCISES. NUMEKICAL PROPERTIES Ex. S. If til.- v.rfti'X aiife-le of an isosceles ti-iangl.) ^40" find tl ^le me tidfd between the altitudes drawn from tlie tiie oppoiiite sitles. Bxtve mitiea o Ex.6. How many degreaa in eac:h angle of an equiu ij:;ula doiieea n? of n equiangular li-gon? Ex.7 How many diagonals are there in a pcntaf-o decagonT inanx-gon! uT i a hexn The methods oi proving that a given angle is a right angle (or that a given line is perpendicular to another given line), or that one angle is the supplement of another, -aiv closely related to tlie ubore tHethuiU of ubtiiinhuj llw iiiiiHi-ri' (■(tl v(il„.''8 of (jivi'H iiiiglef<. Ex. 8. .\i.y iniir of adja.'.^iit aiigleH of ;. parallelogram is mipplu- Ex.9. If oneaugleof aparalie a rectangle. Ex. 10. The bisectors of two (, adjacent angles form a riplit angle ograin is a rigli upplementary (ave verpOii- iterior angles o angle the fignre is Ex. 11. The biseetors of two i the same side of a Z.^ form a right angle. Ex. 12. If one of the It'gs (JB) of an isosceles triangle he produced its own length {ISD) and its ex tremity (D) bo joined to the other end of the base (C), the line last drawn {DC) is perpendicnlar to the base. 193. Algebraic method of proving theorems. The proof ci certain properties of a geometric ligiu'ti is ut'teii fin-ilitiitfil ty the use of an tilgehraic si/iiihoJ for un Kiil.iioini aiKjh "i- tm itjik'iiown line of the. fiifKir. (i>id tin' iifit of uit i<jutilion or ether ulyebrak metkud t/J solution. y Google ;)li I'.OOK I. I'LANE (iEUMlITKi' EXERCISES. CftOUP 3 Al,i;EHtiAIO Mirnillll Ex. 1. Find the number o[ degrees in :iii an^le wlii'.'h eqiiiils twice its complement! [Srti, Let ^ = 1116 complement, etc.] Ex; 2. Fiud the number of degrees in nn nugle which equals its supplement? in one whk'h equals oue-third its supplemeut! Ex. 3. The angular space Kbout a point is divided into four angles TObieli are in the ratio 1, 2, 3, 4. Piud the number of degrees iu eaeU angle. [Sl-o. j:}- 2.i+:!i--t-4J^ = 3tiO°, etc.]. Ex, 4. The ougles of a triangle are in the ratio 1 , 2, a ; lind the angles, Ex. 5- Two angles are supplementary and the greater exceeds the less b>- ^O" ; find tlie angles. Ex, 6, Find all llie .angles of a paraUelogram it one of tiiem is double iLiiother angle. Ex. 7. One of the base angles of a triangle is double the other, and the exterior angle at the vortex is 105°. Find the angles of the triangle. Ex. 8, How many sides has a polygon the aum of whose angles is fourteen right angles? [Sl-g. 2(h-2) = U; fimln.] Ex. 9. How many sides has a polygon the sura of whose angles is ten riglit angles! twenty right angles! 720°i Ex. 10. How many sides has an equiangular polygon one of whose angles is seven -fourths of a right angle f Ex. 11. How many aides has a polygon the sum of whose interior angles equals the sum of the esterior angles ! Ex. 12. How many sides has a polygon the sum of whose interior angles equals three times the aum of the exterior angles? Ex, 13. If the base of any triangle be produced in both directions, the sum of the exterior angles thus formed, diminished by the vertex angle, is equal to twi right angles. [Sua. 180°— a-i-lSO'^— ft-tlSO"— a— (i)=,6te.] y Google EXERCISES. AUXILIARY LINES Ea. 14. TliO bisectors of the base angles of a iaoscelea triangle include an angle -which is equii! the exterior angle at the base. [Sm. To prove « = '!>, denote one of the baso A. 2x, etc.] Ex.15 In an i-osccles triangle the altitude upon one of the legs makes an angle with the base which eqaala one-half the lertcx angle. [Suu To proie «^ Wi, show that n = 90°— r, i = lSO° Ex. 16. If the opposite angles of a qiiadri lateral are eqaal, the figrure is a, parallelogram 194. Use of auxiliary lines. laequalities. The demon- stration of a property of a geometrical figure is frequently facilitated by drawing one or more auxiliary lines on tlie figure. For examples of the nse of such lines, see Props. Ill, V, IX, etc., of Book I. Some of tiie principal auxiliary lines ii!>eil on reutiUucar figures are: a line connecting iwo given points; a line through a given point parallel io a (liven Vin< ; a line- through a given point perpendicular to a gicen line; a line making a given angle with a given line; a line produced its own length, etc. EXERCISES. CROUP AfXILlARY LINES 2i£. 1. In the quadriiaterul AB^AD, and 110 = 0); prove [SuG. Draw AC, etc.] Ex, 2. Prove that the nn(;h of an isosceles trupezoid art S' IJI = / II. y Google 04 BOtHi I, I'L.VKK (IKUilETKV Ek. 3. titiile and prove the converse of Ex. '1. ^ El. 4. GivRii J JS jl ClJ ; provR z ?i = Z ,< ^-■>,,^^ Ex. 5. Conversely, Kiveti Z ?^ '^ — "■'^ = Z ,1 + Z c ; prove JH |1 CD. Ex. 6. Tlio median to the hypotenuae at a ri;;!:! tiiaiiglti is oae-hal£ the hypotenuse. Ex. 7. If one acute angle of a right triangle is double the other, the hypotenuse is double the shorter leg. [Suu. Draw the median to the hypotenuse, etc.] Ex. 8. In an isoBeeles triangle, the Slim of the perpendiculars drnwii from any point in the base I to the legs is equal to the altitude upon one of U the legs. la some cases it is useful to <iraiv firo < Ex. 9. Show that the median of a trapezoid equals one-half the siun of Ihe two bases by drawing a line through the miiipoirt of one IfC of the trapezoid, parallel to the other leg and meeting one base aud the other base produced. Ex. 10. Lines joining the midpoints of the sides of a quiidrilatei-ni taken in order form a parallelogram. [Sui!. Draw the diagonals of the quadrilat- eral and use Art. 177.J Ex. 11, If the opposlti sides of a hexagon are equa and one pair of sides (.4B ami jj; CD) are parallel, the opposite anglcB of the he.-tagoi Ex. 12. 'Hteu Allelic, and An=CJi; prov y Google EXERCISES. INEQUALITIES Vii Let the student form a list of tlie principles proved in Book I concerning unequal lines ottd unequal angles. (See Arts. 92, D3. 95, etc.). EXERCISES. CROUP 19 INEQUALITIES Ex, 1. In the triangle ABC let i' be any pnint in the Bide BG; prove All + BC > AP '.- PC. Ex. 2. In the quadrilateral Alscn let F be any point in the side .Wf; prove perimeter of APCD > perimeter o! AfD. Ex 3. In the triangle AISC let 1) he any point in AB nnd F any point in IW. Ftove AB + BC > An + I)l'-+ FC. Ex. 4. The sura of the four sides of a (piadri lateral is g.eattr IIihu the Bum of the (ilagonala. Ei, B. If, from any point within a triangle, liiiea be diaivn to the verfiees, the sum of the hues drawn is greater tbau one-half the sum of the sides of the triangle. [SUQ. Use Art. 92 three times, etc.] Ek. 6. If, from any point, within a triangle, lines be drawn to the vertices, the sum of the lines drawn is less than the Bum of the sides of the triangle. [Sue. Vse Art, 05.] Ex. 7. The median to any side of a triangle leas than half the sum o£ the other two sides. [SiTG. Prodnee the median its own length.] j Ex, 8. If B is the vertex of an isosei'les triangle vlBr, r. produced to the point J), lliAti ia greater than IBIIA. Ex. 9 lu the figure 51. 7],7t(.' > AB; prove llitV yr* /.BVA. [Sifi, Use Art. 108 } Ex. 10 111 the same figure, show that /■'(' < /.'' . Ex. 11. Ill the qimilrilalet-al liV the shortest : prove ,'- .(/((-srej llian I BAD. Ex. 12 Lines are drawn from .i, ff, ('. ?i, Coiirpoinls in a line, to the point f ontside of the line Wlucli anj;lo3 on 1 atu lefiS than angle ACF^ Which angles are greater I y Google 1 '1 1 BOOK T. I'T.AXE GP:OMETI{Y 195. Indirect demonstrations. Loci. Tii Book I tliri'e nnjthnds of indirect proof hiive been used. 1. The reduction to an absurdity {rednctio ad absiiv- dum), that is, the proof that the negative of a given theorem leads to mi ubsuvdity Csee Prop. 5). 2. The method of exclusion, that is, showing that any other statement ihtn the. i/iven theorem cannot he true (see Pi-ops. XV, SVII). This method is a special ease of the preceding, the negative of a given theorem being divided in it into ti^'o parts ■which are separately shown to be impossible. 3. The method of coincidence, that is, proof that a given line coincides idth annthrr line, irlii'-Ji fulfils cvrioin re- quired cmiditions (see Props. XVII, XXIIJ, XXV, etc.). EXERCISES. CROUr J3 INDIRECT, OE NEGATIVE DEJIONSTKATIONS Prave the folloiyiiig by iiu uvlitneX method: Ez. 1. Every point within an angle and not in the liiseetor of the angle is unequally distant from the sides of the an^ic. [Si,"Q. In the given angle take P any point not in llie bisector of the angle. Then, if P is not unequally distant from AG and on, it must be equally distant from them, etc.] Ex. 2. I£ two straight lines are cut by a transversal, maldug the al:ernate interior angles nnequal, tie lines are not parallel. Ex. 3. The line joining the midpoints of two sides of !i triangln is parallel to the thii'd side. [Sua. Through one of the midpoints draw a line |1 to the Ihini side, show that it bisects the seeond side and that the line joiulug the midpoints eoinaides with it. J y Google EXERCISES. LOCI 97 Ex. 4. IE, ffom a point P in a line AB, Jines PC and FD be drawn on opposite sides of AB making the angle AFC equal to tiie angle BPD, PC and PD are in the same Htraiglit line. [SuG. From P draw PQ in the same straight line with PC and show that PD eoineideB with it,] Ex. 5. The hlaectors of two vertical angles are in the same straight line. Ex. 6. In the triangle ABC, Bis any point in the side AB, and Eis any poi Ex. : a given II tho side Proi i that BE and IJC cannot bisect each Whnt Vi-< EXERCISES. CROUP 1-3 LOCI he iiifus of iill points :d a pivf e this. Ex. 2. What is the locus of all points equidistant from two given parallel lines f Prove this. By use of known loci (see Arts. 112-118), prove the following: Ex. 3. The diagonals of a rhombus are perpendicular to each other. [Su«. See Art. 113.] Ex. 4. The median of tin isosceles triangle is perpendicular to the Ex. 5. The line that joins the vertices of two isosceles triangles on the same biiRe is perpendicular to the base. 1 96. General method of obtaming a demonstration of a theorem. Analysis. A due to the sointion of some of the raoi-e difficult theorems is often obtained by proeeeding thus: Assume the proposed theorem as true; observe tchai other felaiion among the parts of the figure must then be true; proceed backit-ard thus, step by step, till the required theorem 1^ found i-o depcwl on some linoirn truth; then, starting with this knoivn truth, reverse the ste/is tiilrn, (iiid thus build up a direct proof of the nqnireil theorem . This method is called solution by analysis. y Google 98 BOOK 1. PLAXE GEOMETRY The following is a simple example of (lie method : Ex. Given AB and AC Hie logs of an ienacoloa triangle and D any point on AD; prove DC greater than DB. Analysis. If DC > DB, I Bis greater than I DCB. Art, 104. Hence subatituting for ZB its equal, lACli, we have Z^C/; is greater than ^DCB, Ax. 8. But we know that lACB > IDCB. As, 7. Hence, Direct Proof (or Synthesis) ZACB is greater than ITlCn, Ai. 7. .-. Z B is greater than IBCE. Ax.. 8. :.DC > DB. Art. ICU, 0- E. D. The first part (aualysis) of the above process is to be purely mental work ou the part of the pupil, in investiga- ting a given theorem; the second part (the direct proof, or synthesis) is to be written out as the required solution. In working the following exercises, this method will be found to be necessary in the solution of only a few of the more difficult theorems. EXERCISES. QFiOUP 14 THEOREMS PROVED BY VAKIOUS METHODS Ex. 1. If two opposite diagonal connecting thei tills diagonal. ;a of a quadrilateral are blaected by the ices, the quadrilateral U bisected by Ex. 2 Perpendipuiars dras triangle to the median to the b a the extremities uf the base of a Ex. 3. If the perpendiculars from the extremities of the base of a triangle to the other two sides are equal. (1) these perpendiculars make «qual angles with the base, (a) the triangle is isosceles. y Google MISCELLANEODS EXEItCIBEH Ex.4. I( tlie lines Jfi and CD iulorse rsect.tbon Ali+Cn > AC'+DB. iviAes trLacgle is 44^, and one ot praiiuced to me el the opposite Ek, 5. The verlPK aii)|l« cf an u the base anplea is bisected by a hn side. Find ad the angles of the figu Kx,6. In the figure of Prop, V prove that ZJPC is greater than I ABC. Also prove the same m another wiiy by means of an auxiliary line drawn througli fi and F, Ex. 7. Ptrpeudioiilnrs drown from the mi.lpoinl of the ba.^ie of a:a iaosoeles triangle to the leRs are etinal. Ex. 8 State and prove the Ex. 9. In an Isosceles triat Ex 10. In a re ent angle at the re entrant t quadrilateral the exterior tes equals the sum of the dueedtoD; prove BD > JD > JB, Ex, 12. If irOLn a point in the liisector ul au obliqiic aiiglu formed is a rborabus. Ex 14. It the median nf a triang the triangle is isosceles, Ex. 15. Given JB = Ja lBAC = ilB, and DFX ISC, prove A EFA equilaleral. , Unalvsis. 1{ a fJA-i>equilatei = dO^ Her "sing iLbAC= ilC.\ % din U, lfEA = W°; .■.Z»Kfi = GO°; proof, show that IC ~ 30° by y Google 1 00 BOOK I . I'iANE L"; KOMKTliY Ex. 16. If Ihe ilingooals o£ a quadrjlatoral bisect each other at right angles, what kiad of a figure is tlie quaJrilateval f Prove this. Ex. 17. From the point in which the altitudes drawn to the legs ot an isosceles triauf;le intersect, a line ia drawn to the vertex, Prove tiat this lino bisects the attgle at the vertes. Ex. 18. If from a point within an acuta angle perpendiculars ai'e drawn to the sides o£ tlie angle, the angle formoii by thesa perpendictt- tara is the supplement of the given angle. Ex. 19. Linos joining tlio midpoints of tli aides of a triangle divide the triangle into Ecu. eqaa! triangles. Ex. 20. If, in the parallelogram ABCD, BP = VQ, then AQCF is a parallelogram. [A>-.\LYsiS. If J^CPisa^:?, JP=find II QC. .'. hegin the direct proof by sliow ingthat JP=andi3 || QC] Ex. 21. If the diagonals of a parallelogriim are equal, what kind of a figure is the parallelogram! Prove this. Ex. 22. If the cngle A of the triangle AUG is 50° and the oxterior angle BCD is 1L'0°, which is the largest side in the triangle ! Ex. 23. Two triangles are eqnal if two sides and the median to one of these sides in one triangle are equal, respeistively, to two homologous sides and a median iii the other. Ex. 24. Two isosceles triangles are eqnal if the base and an angle of one are equal to the base and the homologous angle of the otiier. Ex. 25. Two equilateral triangles are equal if an altitude of one IB equal to an altitude of the other. Ex. 26. If two medians of a triangle are eqnal, the triangle ia isosceles. [SuG. On Fig, to Prop. XL VI, taking AD= EC, prove A AOC isos- celes, AAEC=l\dDC, etc, How could this theorem be investigated y Google Miscellaneous exekcises Ex. 27. Prove the sum of the aiif-lea of a 1 right angles by drawing a liue through the ve parallel to the baae. :ri:mglft equal to tw rtes of the ti'iangl Ex. 28. The homologous mediaus o£ two equal. equal triangUa ar Ex. 29. The bifleetora of nn angle of a tri- angle and of the twn erterloi- aiigios at the other A Ex. 30. If JKC is an equilateral triangle nTul AP= BQ= CE, then I'QIt is an equilateral triangle. Ex.31. If the two baae ancles of ft trianglo l)e bisected, and through the point of ii;tei'See- tion of the two bisectors a line be drawn parallel to the base, the part of this line intercepted between, the two sides equals the sum of the aegments of the sides included between the par- allel and the base (i, e., prove PQ=AP+ QC). Ex. 32. Two quadrilaterals are equal, if three sides and the two included angles of one are equal to three aidea and the two included anglea of the other, respective!}'. Ex. 33. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. Ex. 34. Lines joining the midpoints of tho sides of a Vtctangle in order form a rhombus, Ex. 35. Lines joiuing the midpoints of the sides of a rhombus l^orm a rectangle. Ex. 36, The bisectors of th? angles of a paraliclogram form a reetangle. Ex. 37. The bisectors of the angles of a rectangle funn a square. Ex. 88. If lines be drawn through the v parallel to tho diagonals, a parallelogram ii M large as the original quadrilateral. y Google 1>I,ANE GHOMK'I'RV grata to thtt luuipo'iit'i of a, pair of opposite sides trisect; of the pavallelogram . Ex. 40. Oq the liiaKOnal AC ot a paralli4osram AltCD ei) AP and CQ, are marked oil, Prove Bl'HQ a jwi'allelogl'a many pairs of equal friaiiglps does the ligiive i-oiitaiii i Ex. 41. Tlie oppo! Ex, 42. In an isosed™ trapezoid, tiie diagonals are equal. Ex. 43. If the upper base of an isosceles trapezoid equals the eum of the legs, and lines be dran-n from the midpoint of the upper base to the exiremities of the lower bas^e, how mauj' iKoseeles triangles are formed ! Prove this. Ex. 44. The lines joining the midpoints of the sides of an isos- celes trapezoid, taken in order, form a rhombus ov a squary. Ex. 45. The bisectors of the angles of a trapezoid form a quadri- lateral whose opposite angles are supplementaif. y Google Book U THE CIRCLE 197. A circle is a portion of a plane bounded by a curved line, all points of wliiisli are equally distant from a point within <mllod the center. The circumference of a (linUe is the curved line bounding the circle. The term itin^le may also be used for the bounding line, if no ambiguity results. A eii'ple is named by nomiiii; its center, an the eircile O; or by naming two or more points on its cireumfereuce, ag tlio uirtlo AVD. 198. A radius o£ a circle is a straight line drawn from the center to any point on the circumference, as AO. A diameter of a circle is a straight line drawn through the cen- ter and terminated by the circumfe- rence, as BC. 199. An arc is any portion of a circumference, as AG. A semi- circumference is an arc equal to one-half the eireumfei-ence, as BAG. A quadrant is an arc equal to one-fourth of a eireumferenee, B.SBD. 200. A chord is a straight line Joining the extremities of an arc, as EF. y Google 104 BOOK Tr. PLASE (lEOMETKV Every chord subtends two arcs. A miuor arc is the smaller of two ares subtended by a chord. A major arc is the larger of two arcs subtended by a chord. Thus, for the chord EF the minor arc is EPF, and the majnr arc is ETF. Conjugate arcs is a general term for a pair of minor and major aros. If the are subtended by a given chord is mentioned, unless it is otherwise specified, the minor arc is meant. 201. A tangent to a circle is a straight line which, if produced, has but one point in eonunon with the circle, aa MS. Hence, a tangent touches the circumference in one point only. A secant is a straight line which, if produced, intersects the circumference in two points, 'as GH. yis. 3 Fis. * 202, A segment of a circle is a portion of the circle bounded by an arc and its chord, as ABO (Fig. 3). Into how many segments does each chord divide a circle? A semicircle is a segment bounded by a semicircumfer- enee and its diameter. 203. A sector of a circle is a portion of a circle bounded by two radii and the are included by them, aa FOQ (Fis, 3), yGoosle THE cmcLE l05 204. A central angle is an an^lo whose vertex is at the center and wliose sides are radii, as the angle POQ (Fig. 3). An inscribed angle is an angle whose vertex is in the circumference and whose sides are chords, as the angle A/JG' (Fig. 4). An angle inscribed in a segment is an angle whose ver- tex is in the are of the segment and whose sides are chords drawn ft'oni the vertex to the extremities of tlie are. Let the pupil draw a circle, a segment io it, and an angle In- scribed in tlie isi'gnieut. 205. Two circles tangent to each other are circles which are tangent to the same straiglit line at the same point. They are tangent hifernally or externally according as one circle lies entirely within or entirely without the other. See the figures, page 122. Concentric circles are circles which have the same center. Let tlie' pupil draw a pair of concentric circles. 206. A polygon inscribed in a circle is a polygon all of whose vertices lie in the circumference of the circle, as ABODE (Fig. 4). A circle circumscribed about a polygon is a cii'ele whose circumference passes through every vertex of the polygon. 207. A polygon circumscribed abont a circle is a polygon all of wliose sides are tangent to the circle, as PQlttiT (Fig. 5). A circle inscribed in a polygon is a circle to which all the sides of the polygon are tangent. Coacyclic points arc poiuts lying on the same circum- ference. y Google 1U6 BOOK n. I'LAKE GEOMETRV PROPERTIES OF THE CIRCLE INFEERED IMMEDIATEIY 208. liiidii of lli<: same circle, or of equal circles, are equal. 209. The flianicier of a circle equals fu-ice its radius. 210. DhiDiefera of the same circle, or of equal circles, are equal. 211. If two circles arc equal (i. e., ma}/ be made lo coincide. Art. 13), their radii are equal, and conversely. 212. A diameter of a circle bisects the circle. For, by placing the two parts of the circle so that the diameters coincide and their arcs fait on the same side of the diame- ter, these arcs will coincide (Art. 197). 213. A straight line cannot intersect a circle in more than tivo points. For, if a straight Hue can intersect a circle in three (or more) points, three or more equal lines (radii) can be drawn from the same point (the center) to the straight line. But this is impossible {Art. 111). Proposition I. Theorem 214. -4 diameter of a circle is longer than any other chord. Given AB a diameter, and CD any other chord in the circle 0. To prove AB > CD. yGoosle THE CIRCLE 107 Proof. Draw Uit radii OC and O'D. Then, in the A OCD, OG+OD> CD. (Whyf) Substituting for OC its equal OA, and for 01} its oqud OB. Ax. s. 0A + Oh, or Ali> CD, Q. E. B, ri!Oi'OfilTIOX If. Tbeoukm 215. 7/i tiie same circle, or in equal circles, equal cen- tral amjhs i)i/crcepl cijuul arcs oil the circumference. Given the equal eircios and 0', aud tlie equal central A. AOBaviAA'O'B'. To prove the are AB=are A'B' . Proof. Apply the circle 0' to the circle so that the center 0' coincides with the center 0, and the radius O'A' ?.'ith the radius OA. Then the radius O'B' will fail on the radius OB, iforhjhi/p. lAOB^lA'O'B'j. And B' will fall on B, Art. 20fi. (/or OB = 0'B', being ra<Ui o/i-<pi<i! ®). Hence arc A'B' will coincide with AB, Art. I!i7. (/or uU poiiils of each arc arc iijiiUUslanI from (JiC i-'cnffr). .'. are A'B'= arc AB. Art. 4T. Q. £. D. y Google 108 l;OOK n. I'LANE UEOME'IliV PrOTOSITION III. ThEOEEM (CONVEE5G OF pROP II) 216. In the same drch, or in eguai circles, equal crn-n subtend equal augks at the center. Given the equal (;irek's niul 0', niiil the efjual ares AB and A'B' subteiuliug the central A ami 0'. To prove ZO = ZO'. Proof, Apply the eirele 0' to the equal circle so that the center & colncii^es with the center and the point A' with the point A. Then the point B' will fall on B, (for arc A'B'^arc AB iy hyp.). Hence the radius ffA' will coincide with OA, and radius O'B' with OB, Avi. CA. {betuieeii two points only one straight line can he liAiini). .'. L and £ 0' coincide. .-. 10= 10'. Art. 47. 0, E. D. 217- Cor. In the fame circle, or in equal circles, of two unequal central angles the greater angle intercepts the greater arc, and, conversely, of two unequal ares, the greater arc subtends the greater angle at the center. Bx, Draw a, circle and hi it a segment whicb la less tbau tlie sector hiviug the same arc. Also oue tiiat is greater. A\ea one that y Google THE CIliCLE 109 Proposition IV. Theorem 218. In fhe snme. circle, or hi eqmil circles, equal chords subtend equal ai-cs. Given the equal eirelos; and 0', uud the elioi'd AB = chord A'B'. To prove are AB = arc A'B'. Proof. Draw the radii OA, OB, O'A', O'B'. Theu, in the A AOB and A'O'B', AB^A'B'. (Wbr I) AO=A'0', and BO=^!i'0'. (Why?) .-. A AOB^AA'O'li'. (Why?) .-. Z0= 10'. (Why J) e O, . ;, arc AB = ai-c A'B'. Ari y fv/iio; ®, eqiKil cmibut A intercept equal a 215. Ex. 1. I„ the above f.^'ure, if thord AB=l iu., chord . XB'--=. aDaare^B = Uin., find the lenRth of iire .J'JJ'. Ex. 2, Draw a clrtlt aud mark off a part of it that is both V. ment HDd a seolor. El. 3. If the diatance fvoiu the eentor of a circle t .0 a li greater than the radius, will tha line iiitetWHct tUo ciri:HHi fercuui y Google 10 UOOK II. PLAMi GEOMETllY Piiopoarnux V. TheoiU':m 219, III ill'' s"»f'' rlrrlr, '/r in equal circles, equal arcs >■€ suhlcnded h>j equal eJiurds. Given the. equal circles O and (T, amlara AB^iiva A' B'. To prove oliord AD — eliord .1'/!'. Proof. Draw the radii AO, BO, A'O', B'O'. Tlien, ill tlie A AOB and A'O'B', 10 = Iff, Art. 216. (for are AJi = A' B' , iiiid, in Ilia snmo O, or in equal ®, equal a U-ml equal A at ihe center). Also OA = 0'A', and OB^O'B'. (Why ?) .-. A AOB- A A'O'B'. (Why ?) .-. AB^A'B'. (Why f] A'J! Q. E. B. Ex. 1. In the above figure if ara 4B = 11 in,, arc ord Ali^l iu., find chord A'B' without meaauring ' = ljlii., and Ex. 2. Draw two circles so that the radius of oc the other. . i. the diameter Ex. 3. To which o£ the elassea of figures mentiousd iu Art. 16 oes a sector ijelong f a segment i y Google the circle 111 Proposition VI. Theorem 220. In the same circle, or in equal circles, the greater of iv:o {minor) arcs is subtended by the greater chord; and. Conversely, the greater of two chords subtends the greater (minor) arc. Given the equal circles and O', and are AB > aro 1>F. To prove chord AB > chord DF, Proof. Draw the radii OA, OB, O'D, O'F. Then, in the A AOB and DO'F, OA=0'D, and OB^O'F. (Why?) Z is greater than Z 0', Art, 217. (for arc, AB > arc I)F, and, in the same G, or in equal 0, of two un- equal arcs the greater are subtends lite greater angle at the center) . .■. chord AB > chord DF, Art. lOT. (if !«■(> A have two sides of one eqitat to tiro siiles of the olhci-, but the included angle of the first greater, etc.). CoNVKRSELY. Given the eqiial eh-eles and (y , and chord AB > chord DP, To prove arc AB > are DF. Proof. In the A AOB and DO'F, OA = 0'D, and OB^O'F. (Why!) AB> BF. (Why?) .•. Z is gi-eater tlian Z 0'. (wiiy ?) .-. arc AB > are BF, Art. 2IT. (in the name O, or in ei/Ma! ®, of tiro viirqual central i llie grimier "™"" '"^"' e, £. B. tnyle intercepts the greater y Google 112 liOOK a. I'LASE (iKOME'lMlY Proposition VII. TiiEOREJi 221. ,1 iiiameter perpendiculnr to a chord hi-wcls the cJwrd and the arcs subtended by the chord. Given tbe circle 0, and the diameter PQ X ch. To prove that PQ bisects the chord AB and the ai and -1 <JB. Proof. Draw the radii 0-4, and OB. Then, in tlic rt. A OAR and OBR, OA = OB. OB = OB. :. AOAB = A OBB. :. AE=BR. and ZAOB = I BOB. :. are .1 P. = arc HP, (in ihr .wmr O, n;- in = ©, = central A intercept = i-d AR. isAPB (Why?) (Why?) (Why?) (Why!) Art. 215. ■<■/.)■ ^AOQ ::=: ^ BOQ {-'^A 15). .\ HTQ AQ ^ ATC BQ. (Wbj t) Q.E. D. 222. Con. 1. -i diameter which l/isecis a chord {shorter than a diameter) i« perpendicular to the chord. 223. Cor. 2. The perpendicular bisector of a chord passes through the center of the circle, and bisects the arcs subtended hy the chord. 224. Cor. '6. A line from the center perpendicular to a chord bisects the chord. 225. Cor. 4. A line passing through the midpoints of a chord and Us arc passes through the center, and is a diame • ter perpendicular to the chord. y Google THE CIRCLE 113 Proposition VIII. Theorem 226. Ill the same circle, or in equal circles, equal chords are eqiudisiaitl from the center; and, conversely, chords which are equidisiani from the center _ are equal. Givea the cii-ele in wliich the eliords AB ami CD are equal. To prove that AB and CT) tire eqnidistant from the center. Proof. Let OE he drawn ± AB, and Of ± CD. Draw the radii OA nnd Of. Then OE bisects AB, and OF Wsects CD, Art. 224. (a Jiiicfriym the milcr ± n chord biser.ls the chord). Hence, in the rt. A OAE and OCF, OA = 00. (Why?) AE = GF. Ak. s. .-. AOAE ^ A OCF. (wiiy?i .-, OE = OF. (Whj?) Conversely. Given circle 0, and AB and CD equidis- tant from tlie (tenter. To prove AB=CD. Proof. Let the pupil supply thu proof. y Google 114 POOK II. l'LA>JE GEOMETRY Proposition IX, Theorem 227. In the same circle, or in equal circles, if two chords are unequal, they are uHequally distant from the center, and the less chord is at the greater distance from the center. Given in the circle the chord CD < chord AB. To prove that chord CD is at a greater distance from the center than chord AB. Proof. Let OG be drawn 1 CD, and OF X AB. Then chord AB > chord CD. Hyp. .-. arc AB > arc CD, Art. 220, (iH the sameO, or in eijml. ®, Ihe greater of two minor arcs is svbtendiil h}i Ihe ijrcat'.T clwrd, mid coiwcrscly) . Mark off on the arc AB the arc AE^&n: CD, and draw the chord AE. Chord AE = chord CD, Art. 219. (t» the same O, of in equal ® , equal arcs are subt/ntiled by equal cltords). Let OS he drawn ± AE, and intersecting AB at L. .: OS = OG, Art. 226. (in the same O, or in equal ®, equal chords are equidistant from the But 0S> OL. Ai. 7. Also 0L> OF. (Why!] Much naore then OS, or its equal OG > OF. Ax. 12. Q. B D. y Google THE CIRCLE 115 pRorosiTiON X. Theoeem (Converse op Peop. IX) 228. Jn the same circle., or in equal circles, if iwo chords are nneqimlly distant from the center, the more remote is the less. Given in t!iR ciriilc llie olioi-d VI> farther from the center thao tlio chord All. To prove ehunl CD < cJioiil J /.'. Proof. Let OH be (iiwwn 1 CTi, mul OG ± AB. OH > CXr. (Why?) Oil OH mark off <>L = 00. Through L let the eliord ELF he drawn X OE. Then chord EF = chord AB, Art. 22G. (in Hie same ,C But the are CD < are EF. As. 7. .-. chord CD < ehord EF, or its equal AB, Art. 220. (in (fte same Q, or i« fqnn/ ®, ilie greater of tico minor arcs is sublended by ilic greater chord, and conversely-), Q. £. 9. y Google lit; BOOK IT. TLAXE GE03IETK1' PliOI'OSlTION XI. TUEOKEJI 229. A KiraUjht line pcrfeiuUcular to a nidiits at Us fxtreinitii is Unnjint to tiic circle. Given the circle 0, tlie i-adiua OA, and \hc line HC ± OA at its extremity A. To prove that BC is tangent to the . Proof. Take P, any point on the line BC except A, and draw OP. Then 0/' > OA. (Why!) Hence the point P lies witliont llie circle. .'. every point in the line BC, except A, lies ontside the O. ,■. BGig, tangent to the circlf, Ari. 2Ci. (« (ann^^il toaO is a straight hi,,- ichkk, etc.), 0. E. B, 230. Cor. 1. The radius drami to the point of contact is perpendicular io a tangent to a circle. 231. Cor, 2. A perpendicular to a tangent at the point of cantad passes through the center of the circle. 232. Cor, 3. The ■perpendicular drawn from the center of a circle to a tungent passes through (he point of contact. y Google THE CIRCLE 117 pROPOSiTioK xn, Theorem 233. Two parallel lines intercept equal arcs on a eir- Case I. Given AB (Fi<r. 1) fcangeat to tlie G I'CD at P, CD it sGiJiuit II AB iiiid iuterseGtiug the circumfciuiieo in C andi>. To prove arc PC = are PD. Proof. Draw the diameter PQ. Then PQ ± AB, AH. 230. PQ 1. CD. Art. 123. .-. are PC = are PD, Art. 221. (a diameter X chord iisR./s (/ib diw.f iKiiJ the ai-e« stMendea by the chord). Case II. Given .iB and CD {Fig. 2) [| seciiiits inter- secting the circumference in A, B and C, D respeefcivoly, Tb prove arc AQ — are BD. Proof. Lota tiLiigent ^i^ he drawn |UIB unLltouehing tlie circle at P. Then EF\\ CD. (Wliy?) Then arc AP = arc BP. Case I, :^ CP= arc DP, CVVhy!) Also Hence arc AC - iBIJ. yGoosle 118 IKJOK II. rLAXK CEOMETRY Case III. Given Ali and CI) (Fig. 3) \\ t;uigeiits toucli. iag the O at P and Q respectively. To prove arc PEQ = are I'FQ. Proof. Let the tmpil supply the proof. 234. Cor. The straight line whk-li joins the points of contact of two 2}ctraUel tangmiix is a dinmeier. Peoposition XTIT. Theorem 285. Through three points, not in the sa %e circumference, and only one, can be ih-a Given A, B am! C any three points not in the same straight line. To prove that one eirenmference, and only one, can be drawn through A, B and C. Proof. Draw the straight lines .4^ and BC, and let _b be erected at the midpoints, D and E, of AB and BG respectively. These -li will intersect at some point 0, Aft. vi'i. {lines ± nmi-paralUl lines are not || ). But is in the X bisector of AB. Const. y Google THE CIRCI,E 119 .*. is equidistant from the points A aud B. Act. H3. In like manner, is in tlio J. bisector of BC, and is equidistant from the points B and C. Hence ia equidistant from the three points A, B and G Henee if a circumference be described with as a cen- ter and OA as a radius, it will pass through .4., B and 0. Also DO and EO intersect in but one point. Art. e4. Hence there is but one center. Again, is equally distant from the points A, B and C; henee there is but ojte radius. With only one center and only one nuliua, but one (lir- cumference can be described. Hence one circumference, and only one, can be drawn through the points A, B and C. Q. E. B. 236. Note. The theorem of Art. 235 enables ua to shrink or economize a circle into three poiiits ; or to expand any three points into a circle. Ex. 1. How many eircurafereneea can be passed through fouf given points in a pliine, each cireumferenee passing tbrou;i:h three, and only three, of the given points I Ex. 2. Draw two eirclea so that they outi have a common chord, Ex. 3. Can two circles whii-h are tangent to each other have a. common chord ! Ex. 4, Can two circles which are tangent to each other have s, common secant 1 Ex. 5. Draw two circles which can have neither a nor a common tangent. Ex. 6. Is it poBsibie to draw two civulus which common secant f y Google I'LANE GKOMET: Proposition XIV. Theorem 237. The two iangenis drawn io a circle from outside the. circle are cqiinl, a>id make, equal (o/y/cs line drawn from the point to the center. point icith a Given PA and PB two tangents drawn from the point P to the circle 0. To prove PA^PB, and ZAPO= ZBPO. Proof. Let the pupil supply the proof. 238. Dbf. The line of centers of two circles is the line joining their eentei's, 239. DBF. Two circles which do not m(!ct m»y have four common tangents. A commoa internal taageat of two circle.'^ is a tangent which cuts their line of centers. 240. Dep. a common external tangent of iwo circles is 3. tangent which does not cut their line of centers. Kx. 1, In the above Sgure, prove that the line ilrawu to the center from the point in which the two tangents meet makes equal aiiglsB with the radii to the points of eontaet. Ex. 2. Draw a circle with a radius of 3 ia. and anothei' with a radius of 3 in., with their oeutera 4 in. apart. "Will these eirelea intersect T n their centers were 6 in. apart, would they intersect S y Google THF, CIRCLE l^J Proposition XV. Theorem 241. Iftico circles intersect, their line of centers is per- pendicitlaf to their common chord at its middle point. Given the circles ami 0', itilevscetinK at tlm points L and B. To prove 00' ± AB at its middle point. Proof, Draw the radii OA, OB, O'A, O'B. Theu OA^OB, und 0'A = 0'B. (Why?) Hence and 0' are two points each equidistant fi'oin 1 and B. :. 00' is tlie ± bisector of AB. ah. in. Braw two cirules in which tlie Una of e Ex. 2. Equals the sum of the lailii. L and />. .-. 00' is tlic X bisector of AB. Avt. 113. Q. E. D. Draw two eireles in which tha 1 y Google V22 liOOK ir. I'LANK (lEOMETKY PiiOPOhiiTioN XVI. Theorem 242. Jf two circles are iaiujeut to each oiker, fke Uite of eenlers passes (hrough the point of contact. Given tho cirdes iiml 0' tangfiiit to each otliev at tbe poiut R. To prove that the line 00' passes through It. Proof. At the point E let PQ, a tangent to the given ®, be drawn. Also let AB be drawn 1 PQ at R. Then AB passes through and also throngh 0', Art. 231, (a l-loa tangeni al the fioint of coiilact pusses Ihrough the center). :. line AB eoincides with the line 00'. Art. 64. .'. 00' passes through the point E, (Jor it coiittidm ititli AB ahieh passes through li ) . How many eomnioil interail, and tangents bave two eirolea Ex. 1. If they toneh esternally t Ex. 2. If they touch iDternally f E*. 3. It they intersect ! Ex. 4. If one circle lies wholly with Ei. 5, If one circle lies wholly witl external y Google THE CIRCLE eXERCISES. CROUP I Ex. 1. The line joining the oenlor cKord is perpeadienlar to the chord. I four points t alt en i 'O is tho perpendieul 1' upon two clioi'ds a [■0 BD; prove that chord Ex. 2. A, B, Cand D n. eircumferenee of a eirele, AC=ahovii BD. Ex. 3. Taugeuts drawn at the estron p&ratlel. Ei. 4. PA and FB aro tangents to a eii P. is the center of the oirele. Prove th; bisector o£ the chord AB. Ei. 5, If the perpendiculars from the Ci equal, tlio aroB subtended by these chords a Kl. 6. A, B. C and D are points taken i onmfecence, and arc AC is greatei' than t AB> chord CD. Ex. 7. State the converse of the preceding theorem and prove it. Ex. 8. Given EA, EQ and QB tangents o£ the circle ; prove Rq=EA + QB. Ex. 9. If a quadrilateral be e ire urns cribed about a. eirele, show that the eum of one pair of opposite aides equals the auiu of the other pair. Ex. 10. if a hexagon be eireumscribedabout aairolo, show that tho aiim of three alternate aides equals the sum oC the other throe sides. Ex.11. If a polygon of 2rt sides (he sum of n alternate sides equals t: Ex. 12 Acm inequilateral, Ex. 13, If two circles b externally, the eoiuiuuii intei bisects the common external ti prove PA = FB). ribed paralleiograi mgent .tl,i., yGoosle il;4 BOOK II. I'L.VNR nKOMF.TliV Ex. 14. It two circles are tniigfi (either e.tteniiilly or iiitei-nally) taugcti drawn to llieiu from any point iu tie eon mon tangent ara equnl. Ex. 15. Two eircli O' are tangent mternnJly at P. The line PAB ia drawn intersecting the Prove that OA and O'B MEASUREMENT. RATIO. 243. Measurement. For many purposes, the most; advantageous way of ilealiog with a given magnitude is to take a certain definite part of the magnitude as a unit, and to determine the nnmbei' of times this unit must be taken in order to make up the given magnitude. Ease and precision iu dealing with magnitudes are thus obtained. Geometric magnitudes thus far have been treated aa wholes, the object being simply to determine whether two given magnitudes are equal, or unequal, or to determine some similar general relation. Hereafter geometric magnitudes will frequently be treated as if com- posed of units. To measure a given magnitude is to iind how many times the given magnitude contains another magnitude of tlie same kind taken as a unit. 244. Tlie numerical measure of a magtiituJe is the num- ber which expresses how many times the unit of measure is contained in the given magnitude. Thns, when a boy says that he is five feet tall, he means that, if afoot rule be applieti to Lis height, the foot ruie will be contained live times. A quantity is often measured to beat advautage by measuring a related but more accessible quantity, which Las the same numerical meas- ure as the original quantity. This process is indirect measurement. Thus, the temperature of the air is measured iniiirettly by measuring the height of a column of mercury in a thermometer tube. So the number of times a unit of angle ia contained in a given angle is often ascertained y Google MEASUKKMKKT 125 moat readily by determining tbe number of times a unit of arc is cod- taiued ttt a give □ a. re. 245. The ratio of two niiigniUKius of the same kind is their relative magiiitatle as determined by the number of times one is contaioed in tbe other. Hence, it is the qiio- tient, or indicated quotient, of tlie two miignitudos. Thus, tbe ratio of 3 ft. to I ft. 7 in. ia ^~. <"■ fg- Tlie use of ratio U illustrated by tUe fuct tbat several indicated quo- tiimts wbcil taken togetlier may be simplified by canoellation before a floal determination of tbeir vnlins La iiinde. Two magnitudes of the same kind have tlie same ratio as their numerical measures. 246. Measurement as a ratio. Au important piirticular instance of ratio is tbat ratio in wliicb one of tbe two mag- nitudes compared is a unit of measurement. Hence, the nnmerical measure of a magnitude is the ratio of tbe mag- nitude to tlie unit of measure. Tims, tlie numerical meaaure of tbe beiglit of a boy is tbe ratio of, bisIieigU(5 ft.) to tbe uait of measiiru (I ft.), orS. 247. Commensurable maguitudes are magnitudes of the same kiud which have a common unit of measure, Tiius, 12 ft. and 25 ft. bave tbe common unit of measure, 1 ft., and ience are commensurable magnitudes ; also 13! bu3, and 7J biia, liava a common unit, I peclc, and are commensurable, 248. Incommensurable magnitudes are magnitudes of the same kind wliich have no common unit of measure. Ills, $5 and Iv'a ; 7 vrs. and -^15 yrs.; so tbe side and the diagonal of a squHve may be proved to bave no common unit of measure ; likewise tbe diameter and tlie circumference of a circle. In general, a ratio wliicb is expressed bj a surd number, ^\/S or\ 3, is a ratio between incommensurable magnitudes (called au incommensurable ratio). y Google 'I,ANE GEOMETRY METHOD OF LIMITS a variable, is fi qnanlity imlK'f of differfijit valnes or |3voi)ltMii . goes varies with the nurabei of tou w h 250 \ constant u ivhicli remains iincliiiiigeLl in \ u a Lissioii. h g circle and tlie iiuinUcr of sides oftb itc, the pariiiietet of tUe pnly- j;OI\ e 1) iioe of tlie circle nill remain uncLan^ an li b u 251 T mto JJ mtity is a conatiiut quantity ^fiven discussion, the given \-M 5 we please in Talue, but 1 the circumference of tlio circle ]a til lie n e bed i>olygun ; also,- the area of fUe circle is the limit of the urea of the inscribed polygon. From the definition of limit, it follows that the differ- finee between a variablo and its limit may be made as small as we please but can not become zero. As another illustration of a variable and its limit, we may talie the ease of a point Ptravelling along a given line AB, in auch ft way that in the first second it passes over APi, one-half the line AB; in the second second over half the remaining part o£ the line, and arrives it P;] in tlia third second over one- half the remainder of the line, and , , , , arrives at Pj, etc. It is evident -^ fi ^' ^^ ^ that the point P can never arrive at B; for, in order to do this, in some one second the point would need to pass over the whole of the remaining dietanee. In this illustration, AP, the distance traveled by the moving point, is a variable (depending on the number of aeconds), and AB is its limit. If the distance AB be denoted by 2, the distance traveled will be denoted by 1 -fl + i -|- i + , . . . and will vary according to the number of terma o! the series taken. y Google SfETHOD OF LIMIT3 127 252. Use of variables aad limits. Many of the proper- ties of limits are the same as tlie properties of the variables approaching them. Ileuce a demonstration of a difticiilfc theorem may often bo obtained by first fintling the properties of relatively simple variables and then transferring these properties to their more complex limits. 253. Properties of variables and limits. 1. The liinit of iJte sum of a nitmbfr of eari^lfS equals the sum of the limits of these vaHables. For, since the difference between each variable and its liinit may be made as small as we please, the sum of alt these differences may be made as small as we please (since it is a finite number of differences, with each difference approaching zero). 2. The litnit of a times a variable equals a iiiiies the limit of the variable, a being a constant. For, if the differ- ence between a variable and its limit may be made as sinall as we please, a times this difference may be made as small as we please. 3. The limit of -tk part of a variable *■'< -fh part of the limit of the variable, a being a constant. For, if the difference between a viiriable and its limit may be made an small as we please, — th part of tliis difference may be made aa email as we please. 4. If a variable ~0, a times (a being finite) or - part of the variable ~0; and if A diminish in the one, or i the other process, the limit is sliU zero. Ez, 1. Are 2) gal, mid 3^ qt. c Ei. 2. If c donotc u miistaiit bi eacU ot the following a v&riable or a Uftte. y Google LASE (iEOJlETRY XVII. Tin:f)iu.:M 254. 7/ tu:> rariabU^ nr- .iJmnj^ pqml, and mch < .■or.clu^.i a UntU, /luir Umil^ un- 'qiuil. y I> <- GiTen AB flic limit of the variable Al', AC the Ihuit of the variable AQ, and AP^AQ always. To prove AB = At7. Proof. If the limit AB does not equal the limit AC, one of these limits, as AG, must be larger tiiau the other. Then, on AC, take AD equal to AJi. But AQ may have a value greater than AD. Art. U.jl, Hence AQ would be greater than .i/>, Or AQ > AD tt-hi.)h = AB) . . AQ > AP, {Jor AP). But this is eontrary to the hypothesis that AQ aud AP are always equal. ,'. AC cannot be greater than AB. Id like manner it may be shown that AB is not greater than AC :. AB^AC. Ex, "Wliiit method of proof is used in Prop, XVII! y Google MEASUREMENT OP ANGLES Proposition XVIII. Theorem 255. In the same circle, or in equal circles, ix-o angles have the same ratio as their intercepted arcs. Case I. T17jfw the intercepted a tral 'mmensuraile. To prove Given thuoqiuil® C and 0, with tlie ceutrai ^A'O'B' ami AOB intercepting the commensurable arcs A'B' and AB. ZA'O'B' ^A'B' lAOB AB' Proof. Let arc PQ {from n circle - ami 0') be ix com- mon moasiiro o£ tho arcs A'B' and AB. PQ will be contained in arc A'B' an exant number of times, as 7 times, and in AB an exact number of times, as 5 times. arc AB 5 From (y and draw radii to the several points of division of the arcs A'B' and AB. Then the ZA'O'B' will be divided into 7, and the /.AOB into 5 small angles, all equal. Art. 216. {in the «!. Fj =® equal arcs mbtend eqviU A. at the t lA'O'B' ^1 lAOB 5 ZA'O'B' ^A'Ii' I AOB AB' Art. 245. y Google BOOR II. I'LAXE Cask II. inint the h<lr>rrpf''d an siirahle. Giventheeriiiii! ciiO'iimlO,witli the eeiiti'id A A'O'B' a.hA AOB iuteruepting the iuooramensorable arus A'B' and AJi. To prove ZA'O'B' _ ' lAOli AlV AH ' Proof. Let tlie arc AB be divided into any naniber of equal parts, and let ooe of these parts be applied to the arc A'B'. It will be contained in A'B' a certain number of times, with an arc BB' as a remainder. Henoe the arcs A'D and AB have a common unit of mess- lA'O'D A'D lAOl) AB ' Case I. If now wu let tlie unit of measure be iiidcflii ilelydimin- ished, the arc DB', which is less thun tlie unit of i ncasuro.will be indefinitely diminished. Hence arc ^'i>=^arc A'B' a-s a limit, and I A'O'D- ^1^'0'5'asalimit. An 351. Hence , . ..„ becomes a vafiable approaohi ZA'O'B' " lAOB as its limit; Alt. 253,3. Also -j^ becomes a variable approaching A'B' . limit. Art. 253, 8. y Google MKASri!EMENT OF ANGLER 131 Lsel. 256. Dep. a degree of arc is one tliree luiiiJf.'d and sixtieth part of the eircura fere nee of h (circle. 257, Cor. TJie, mcniber of degrees in a cfnttul umjle equals the niimher of degrees in /he hitercepled an: ; thut is, a central angle is measiirrd by Ua Intercep/ed are. Ex. 1. What is the ratio ot a quadrant to a semi-o Ira u inference ! Ex. 2. "What is the ratio of an angle of an equilateral triangle to ne of the acute angles of an isoaeeles riglit triangle t ot each circle is on the Ex. 4. Draw three circles so that the center o£ each ia on camterenee of the other two. [SuQ. First draw an equilateral triangle.] Ek. B. Draw three circles each of which shall he tangei other two. Ex. 6, Draw tv to one ot these cirt Ex. 7. Draw two concent rit one and a chord o£ the other. y Google liOOK II. PLANE GEOME Proposition XIX. Tbeork.u 258. An inscribed angle, is measured by one half ifg Inlercepted arc- ana AB (Whyf (Whyfl Case I, When the n-nl the inscribed angle. Given Z2J.4C(Fig. 1) iusei-ibecl iii the I'ircle C passing through the center 0. To prove that Z BAC is measured by * aw. BC Proof. Draw the radius OC. Then, in the A OAC, OA = OC. I A = IG. But lIiOC= ^A +Z0. .-. ZB0C = 2Z.-1. But Z HOC is measured by arc RC, {aceitlral Z iiineo-mredhmlsixlcrccincd arc). .'. Ko. of iingulur degrees in / .BOC^No. of arc dfgi Hence Z -i is iiieusnred by \ arc SC. Ai. 5. Case II. Wheii the center of the circle lies within the ht scribed tmgle. Given the inscribed /IBAG (Fig. 2), with the eentec of the circle tying within the angle. To prove that /.BAC is measured by h arc BO. Proof. Draw the diameter AD, Then /.BAD is measured by ^ are BD. Caael. Let the pupii complete the proof. aBC. yGoosle MEASUREMENT OF ANGLES 133 Case III. When (he cenier of the circle is witliout the inscribed angle. Given the inscribed IBAG (Pig. 3) with tlie center outside the angle. To prove that ABAC is measured by h arc BC. Proof. Let the pupil supply the proof, 259. Note. By use of the nbove theorem, if the number of degree* in the intereepfed arc be known, the number of degrees in the insoribed Bnslo enn be determined immediately. Thus, if the arc BC (Pig. 3) eoQtaixiS 4S°, the angle BAC contama 24°; also, if it be known that the mslo BJC e< 260. CoE. 1. AU angles inscribed in the same segment, or in equal scgmeni.t, are equal. Thns A A, A', A" (Fig. 4) are all equal, for each ot them is measured by one -half the arc BBC. 261. Cob. 2. An angle inscribed in a semicircle is a right angle, for it is measured by one-half a semicircum- ferenee. Thus FE (Fig. 5) is a diameter .-. ZFGEis a rt. Z. 262. Cor. 3. ^1k angle inscribed in a segment greater fhan a semicircle is an acute angle; an angle inscribed in a Segment Jess than a semicirch is an ohttise angle. y Google iS-l BOOK II. PLANE CEOMI'mji: Proposition XX. Theoreji 263. An angle formed bi/ two diordn Intersecting leiihin a cireiimfiirence is medsurrd hij half the sum of the inter- cepted arcs. Given tlic chords AB and CD in the G AVBC, inter- secting within the circiimferenco at''tlie point P. To prove that /.DFB is measured by J (arc BB-\- ai-o AC). Proof. Draw the i^hord AB. Then, in A ABB, IDBB= IA+ IT), Art. 13S. [an exi. I of a £\ is equal lo the sum of the lu'o opp. int. i ). But Z J. is measured by i are DB, Art. 358. {an inscribed Z is measured l/ij one-half its mlcrvepte/i are). Also Z 7> is measured by i are AC. (Why!) Hence I DFB is measured by A (arc JJB + arc AC). Ax. 3. Q- E. B. Ex. 1. lu the above figure, i! arc />/( eonUins W and are -IC coa- t*ms 38°, how many degrees in lAPC * ia lAPD ? Eic. 2. If arc AD = 5i;° and Z AFD = 124°, fiud are CB. Ex, 3. If, in Fig. I, p. 132, arc AC oofttalna 112°, how many degrees are there in tiie I A ! y Google MEASUREMENT OF ANGLES 135 Pedposition XXI. Theorem 264. An angle formed by a tangent and a chord drawn from the point of contact is mfamred by half the inter' cepied arc. Given the O PCD, awl Z.APC formed by the tangent APB and the chord PC. To prove that Z APC is measured by i are PEC. Proof. Let the ehord CD be drawn ll AB. Thtii Z.APC -^ ■- I PCD. (Why?) But I PCI) is maasu red by I a .i-c PD. (Why!) .-. I APC is measu red by i a re PD. Ax. 8. But (i. arc PEC = 00 W Uiie,^ inkrcept cijuw arc DP, ■rcumfermce). Art. 233. ,■, /APC is measured by i arc PEC. Ax, 8, Also Z.BPC, the supplenieut of ^APC, is measured by i are PDC, Ax. 3. (for urc I-DC is the conjmatc of arc I'liC) . Q. E, B. Ex. 1. In the iibove figure, if are /'A'(,* uoKtaius l::4'^, Low many degrBea lire in the angle AI'C ! Ei. 2. If aru CD = 'M", find the angles on the figure. y Google rKorosmoN XXJT. Thkorrjc 265. An an'jle formed by two secants, or hij lirn tangents, or hi! a secant ami a tangent vieel in;j withmil the circumference is measured by half the ilifferntre nf Hip hilerrrjiled ores. FiS.t 3?is.%. (fib', a I. Given the O ACDB (Fi^. 1), and the lAPB formed hy the two secants PA and P£, meeting at the point P with- out the circumference. To prove that ZP is raeasnrerl hy 1 (arc>4B— arc CD). Proof. Draw tlie chord CB. Then IACB= ZP+ZB- [Whj-i) .■. ZP= ZACB— ZB. Ax.3, But ZACB is measured by i arc AB. Art. 258. Also IB is measured by 4 arc CD. (Why!) .•. /P is measured by A (arc ^P — arc CD) . Ax.s. II. Given the O ABB (Fig. 2), and ZCPB formed by the tangent PC and the secant PP. To prove that Z P is measured by * (arc ABB — Are AD). Proof. Let the pupil supply the proof. y Google KXKliCTSF.S OK THE CTRCLK 137 m. Given Z^Pfi (Fig. 3) formed by tlie tangents PC and PD. To prove that ZP is measured hy i {arc AEB — are AFB) . Proof. Let tho i>iipil supply the proof. 266. Note. By means of Props. XVIII-XXII, angles formed by chords, seeanta, or tangents, or oombinations of these, are all reduced to eentrp.1 angles and hence may readily be compared. Ex. 1. If, in Tig. 1, p, 130, are CI> = Zi° and arc .JJV = lOS", draw AD and find all the angles of the figure. Ex. 2. If, in Fii-. 2, p. 13U, angle i' = SO'', find arcs-Ji'fl and AEB. EXERCISES. CROUP tS Ex. 1. What now methods of proriug two lines equal are afforded by Book ir t Ex. 2. Wliat new methods of proving two angles equal '. of prov- ing two angk-5 supplementary f of proving an angle a right angle 1 Ex, 3. What methods of proving two arcs equal ! Ex. 4. In ii quiidrilareral ioscribed in a circle, tiicli pair of uppiwiti; angles is supplcinciitiiry. Ex. 5. A «hord forms equal angle? with the tangents at ita extremities , Ex. 6. If an isosceles triangle be inscribed in a circle, the tangent at its vertex makes equal angles with two of its sides and is parallel to the third side. (Is the converse of this theorem inic ?) Ex. 7. I£ two chords in a circle intersect within the i^irele at right angles, the sum of a pair of alternate area equals a semioircumf erenee . Ex. 8. Given the center of a circle and -iC a tangent ; prove IBAC=X<<^0. Ex. 9. If one side of an inseribed quadri- lateral be produced, the exterior angle fio formed eiiuals llio opposile interior hi y Google i;;s ];ooK II. ]']^ANE ^.EO^rKTuy Ex. 10. IE twi> tiingoiits (o a. oirele incluiio an nn^iia of 60" at their point of intersection, the chord joining the ]ioiut3 of eontaot forms with the tnngenta an equilateral tl'iangle. Ex. 11. Tliu chord A/I equals the eijord C]) lu a given eirele, and the chords, if prodneoii, iutei'svet at the point i'. Prove seeant PA^s secant I'C pribed ahout the triani^le ABC, and P ia Prove that the angle Jill' equals oue- Es. 12. A circle iE the uiidpoint of the a half the angle C. Kk. 13. It A, n, C, 1> and E be poiTits taken ii eireumtercnee of a eirele, and the area AU, BC, CIJ and /IE be equal, prove that the angles AliC, HCI) and CDl! are equal. Ex. 14. An inscribed angle formed by a diameter and a chord baa its intercepted are biaeeted by a radius which ia ,, parallel to the chord. Ex. 15. Two see^mts, PAI! and PCD, intersect the eirele ABDC. Prove that the trian>;les P/IC and rj/>are mutnally equiangular. Ei. 16. Given AC s, tangent ami An\\ri:,- prove A ACD and AJIE rautnally equiangular. ' Ek. 17. Given ABC an inaoribed triangle, AE ± BC, and CD X AB ; prove arc iJO=arc BE. Ex 18. If the diagonals of an iuseribed quad- rilateral are diameters, the quadrilateral is a Ex. 19. Tangents through the voitie. inacribeii I'eptangle form a rhombus. Ex. 20. Two circles intersect at /' and Q. PA and Pli are diameters. Prove that <^A and QD form a straight line. Ex. 21. Two eqi at Pand Q, through P a line ia dra terminated by the eireumfersooes equals QB. [SuQ. i A and B are measured by what arcs !] J and B. Show that QA y Google EXERCISES. AUXILIARY LIXEK 139 267. Use of auxiliary lines. In deraoustfatiii^ theorems relating to tlie cifcie, it is often helpful to draw one or more of the following; auxiliary Hues: A radius , a diame- ter, a chord, a perpeirdicuhtr from the coikr upon a chord, an arc, a circtimferfiu-i^, etc. EXERCISES. CROUP 17 AC'XTI.IARY LIXKS •J ; prove BM = r.X. Ex. 2. If from any point in the eircuraferenee o! a circle a eliord and a tangent be drawn, llic perpendiculars drawn to tLem from the midpoint of the are subtended by the chord are eq^ual. Ex. 3. Given the center of a eirule, and 0J> X chord AC ; prove IA0D= IB. Ex. 4. Prom tbc extremity of a cHametcr chords are drawn making «iual angles ivitli tiie diameter. Prove that these chords are equal. ISvG. Draw Ji from the center to the chords, et la the converse of this theorem true ? Ex. 5. If through any point within a circle equal chords he drawn, show that the line drawn from (he center to the point of intersection of the chords hisects their angle of intersection . Ex. 6, Tangents /'J and W are drawn to a cir O. Prove that angle i' equals twiue angle OAIl. Ex. 7. If in B the segments of o other chord, Ex. 8. A paru rectangle, CSuo. P, 100.] le Iwp equal chorda ir orii equal the segment ( the diajjouiils o£ thu i—/ , y Google rr,ANF, riT'O.MKTRY Ex. 9. If a qundvLlatera! Ije o ire am scribed nboiit a circle, Um anglea subtended at the center by a pair of opposite aidfis are supplpinentacy, [Sua, Draw radii to tlie poiiifce of contact and show that tiiere are four pairs of equal A at Hie center.] Ex. 10. If an equilateral triangle ABC be inscribed in a oirele and any point P be taken in the arc AB, show that PC= PA+FB. [SUG. On PC take Pif equal to FA, draw AM and prove ii PAR and MJC equal.] Ex. II. Two radii perpendicular to each otlier are produced to intersect a tangent, and from the points of intersection othei' tangents are drawn to the circle. Prove that the tangents last drawn arg parallel . [Sun. Draw radii to the three points of con- A tact, etc.] Ex. 12. A straight line intersects two oonot trie circles. Show that the segments of the I. intereepted between the eirpumferences are eqi (prove --15 = 01)). Ex. 13. A common tangent in drawn to two circles wliich ara exterior to each other. Show that the chords drawn from the points of fangenoy to the points where the line of centers cuts the cir- cumferences are parallel. Ex. 14. A circle is described on the radius of a diameter, and a chord of the larger circle is drawn from tlie point of coataet of the two circles. Prove that this chord i circumference of the smaller circle. [SuG. If the chord is bisected, a X from the c( the chord will also bisect ti:B chord, i Ex. 15. Two circles are tangent ex- ternally at the point P. Tiirough F any two lines APB and CPD are drawn ter- minated by the circumferences. Show that the chords AC and BD are parallel. [Sl'O. Draw the common tangent at F, If AC and BD are |1 , what A mui ■ted hythe of the larger to- y Google EXERCISES. MAXIMA AND MINIMA Ex. 16. Two circles intersert at the points P Biid Q. and CQU uro drawn, terminated by the oiroumferences. ^Cand BDare parallel. Ex. 17. If a square be described on the hypotenui triangle, a line drawn from the center of the square to the right angle bisects the rigrht angle, [SuG. Describe a circumference o triangle as a diameter.] Ex. 18. If two cirelea are tangent tangeut touches them at A and B, re right angle. Ex. 19. If in the triangle ABC th drawn, the augle ASD equals the angle AEB. [Sua. Describe a semieireumference on A righL 1 the hypoter ixternally at !', and : pectively, the angle altitudes JIV and AE a diameter.] 268. Def. a maximum is tlie greatest of a class of mawiiitudes satisfying eertalE given conditions, and a minimum is the least (see Arts. 470, 471) . For instance, of the chords in a given circle, which is the maximum 1 EXERCISES. CROUP 18 MAXDIA AND Ex. 1. 0! the chorda drawn through determine which is the greatest, and all Ex. 2. Find the shortest line, and also the longestline, that can be drawn from a given external point to the eir- eumterenee of a circle. [Sl'G. O beinp the center, prove in A PCO, PA < PC ; by A PDO, PB > PD.] Ex. 3. Find the shortest line, and al est line, that can be drawn to the cii of a circle from a point within the circle. [SuQ. O being the center, prove, by A Ol'C, FB < PC. etc.] y Google 1-i'J liOOK II. PLANE l.!E Ex. 4. If two cirples intersept, show tU»l, of Hties drawn throufrli a piiict of , iJii-'ifei^tifiu and fenainnted by ihe> cir- (tumfereiitt'S, that line is a loaximiim which is pai'allpi to tlm liueof <^e»t6rs. [Sun. Prove US < 00' .: CPD < AI'B .] Ex. 5. Given AB 1 Oli in eirele O ; prove ZO.rBtbe ffiasimuraof all i having their vertieea im the oivcumferecee and their sidespasHing througli O and i! roepeetively. a eirelo on OA as a diameter.] EXERCISES. GROUP rs DE MO XST RATIO XS BY IXDIRECT MKTTIODS Prove the following by an indirect method (see Art. 195) ; Ex. 1. A segment of a circle which contains a right angle ia a semicircle. Ex. 2. If a rectangle be inscribed in a eirelo, its diagonals are Ex. 3. Prove the second part of Prop. VI by an indirect method, Ex. 4. Prove Prop. X by an indirect method, Ex. 5. A straight line connecting the midpoint of a chord and the midpoint of the arc subtended by \\\e chord is perpendicular to the chord (use the method of eoineidenee). Ex. 6. A line joining the midpoints of two parallel chorda passes through the center. [Sl'g. Draw a J- to each chord from the center and show that Ex. 7. If the opposite angles of a quadrilateral are supplementary, a circle oan be circumscribed about the quadrilateral, [SUQ. Pass a circumference through three vertices of the quadri- lateral; it it does not pass through the remaining vertex, etc.] Ex. 8. Prove the converse of Prop. XII, y Google EXERCISES . CONSTAKTS AND LOCI un 269. A geometrical constant is n, gfom(;trteal inagiii- tude wliifih varies in some respect, as in position, but remains constant in size. Thus the angles inscribed in a given semicircle vary in position but are alt of the same size; viz., a right angle (see Art. 261). EXERCISES. CROUP 30 XATIOX OF CON'STAXTf^ AXD LOCI Pis any point Ex. i. All and AC are tangents to a eirck cirouralerenee outaids tlie triangle ABC. As sura of tiie Z.-iand / JiPC is conetant. Ex. 2. In Es, 8, p. 12n, if AR and BQ be produced to meet at T. Bhow that tlie venmeter ot tho triangle Tffy equals tlie sum of TJ and TB, ami heiiee that the perimeter of triangle TBQ is constant, no matter how /' may vary in position betmeiin -J and IS. Es. 3, Shoiv o the ■e that, i£ O i the c , Z.ROQU Ez. 4. Two circles intersect in the points A aod B. Prom any point P on one ciraum fereuee lines FAC and PBD are drawn, ter- minated by the other eireumference. Show that the chord CD is constant. [St-G. Draw BC and prove ICBD a, Ex. 6. Given OA J- OB, and Cn : (liven lungth moving so that D is alwa; and C in OH and 7' the niidjiohit c that OP is conf^faiit in IciikMi {-.c-e V The dftermmatioH of ha fucilitaied hij shoiviiig that magnitude is a conslunt. Es. 6. Find the locus ot a poi dlataiiCL* a troui a giveQ circumtei'i y Google 144 isOOK II . I'LANE r.EOMETUY Ex. 7, Fiml Tho locus of Iho mUipoiiils ot the ni Ex. 8. Find the locus of tho mi.lnoiiits of :in chordsi of n ftiven leugtli di'awiL in a given eirole. Ex. 9. Find tlie )oi>us of tiie vertices of nU rij^lit lriaJii;li'K liaviag a giTen liypotouuse. Ex. 10. Find the loeus of the raidpointa of all tht ohoi'ds drawn [Sl'G. Draw a line from tho center to the given point and perpen- diculars from tbe teirtcr iipou fhn chords.] Ex. 12. QP length and always in a line. Find the EXERCISES. CnoyP 31 THEOREMS PRO^TA) BY VAKIOTJS ^IKTHODS Ex. 1. The lino which liiseeta the angle formed by a tang a ehord biseefs the intercepted are also. Ex, 2. An inscribed trapezoid is isosceles. Ex. 3. Given TA and TI! tangents, are J7.'= T^ m", an Sll = s:,°, and are 1)0=150°; find all the angles of the figure. Ex. 4. If two tangents to a circle are parallel, the line joining their points of contact ia a diameter, [SuG. Draw radii to the points of contact and use an indirect method of proof,] Ex, 5. A rectangle cireiimaerihed about a circle ii [Sua. Use the preceding theorem.] y Google MIKCF.LLANEOUS EXERCISES Bx. 7. Find the angle formed by thi Bide o( an macrlbed square and the taugeii through the vertex of the square. Ex. 8. Prom an external point a seeant is drawn through the Center of a circle, and also two other secants making equal angles frith tbe first secant. Show that the seeanta last drawn are equal, Ex. 9. AliC ia a triangle and on the side AB the point P ii taken and on liC the point Q, so that angle Bl'Q equals angle C. Show that a circle may ba circumscvibed about the quadrilateral Ex. 10. Two oiiClBB are tangent to each other externally, and a line is drawn thtough the point of contact terminated by the circum- ferences. Show that the radii from the extremities of this line are PMallel. Ex. 11. If a Wangle as di^u triangle. Ex. 12. The chord of an are is parallel to the tangent at the midpoint of the are. Ex. 13. I£ a triangle be inscribed in a cirele, the sum n£ the angles inscribed ia the segments exterior to the trianijle is four right auBles. Ex, 14. Find the corresponding theorem for an inscribed "juadrilateral. Ex. 15. Find the loi^us of the centers of all circles passing through two given points. Ex. 16. If two unequal chords intersect in a circle, the greater chord makes the less angle with the diameter through the point of intersection of the chorils. e of this theorem. Is the Ex. 17. The .51 tbe hypotenuse am y Google UG BOOK TI. PLANE GEOJtETKV touch the m- iiflo FQR and Ex. 18. The aidos AIS, BG and AC of u tr Bcribed circle at the poiuts I', Q and U. Show one-balf angle A are complementary. [Suo. Draw radii from the center to P and IE. Then I PQR IFOA, etc.] El. 19. Fi-om the point in whi angle meets the circumforeiieo, a ch of the angle. Show that this chord equals the other aide of the angle. Ex. 20. Given AB a diameter, AP= the radius, AD and PC iBngents; prove A CED equilateral. [St'G, Draw OC and CA; then in rt. A OOP, C^ = radma, ZP^GO", etc.] Ex. 21. Perpendiculars are drawn from the extremities of a diameter upon a tangent. Show that the points in which ihu perpendi the tangent are equidistant from the cente:. CONSTEUCTION PROBLEMS 270. Postulates. As statod in Art. 49, a postulate In geometry is a coustrmitiou of a geometric figure admitted as possible. The postulates used in geometry, aru as follows (see Art. 50) ; 1. Through any two points a straight line may he drawn. 2. A straight line may J>e extended indefinitely, or it may be limited at any given point. 3. A circMmference may he described about any given point as a center and with any given radius. y Google CONSTRUCTION PROBLEMS 147 271. Tlie meaning of the postulates is that only two drawing instruments are to he used in making geometri- cal constructions; viz., the straight-edge niler and the compasses. With these two simplest drawing instruments it is ble to be able to construct as many geometrical 272. Form of solution of a problem. The stiitoment of a problem and of its solution consists of certain distinct parts which it is important to keep iu miiid. These are 1. The general enunciation. 2. The particular enunciation. (1) Given, etc. (2) To construct, etc. (or some other construction phrase, as "to draw," "to bisect," etc.). 3. The construction. 4. The assertion. 5. The proof of the assertion. 6. The conclusion (indicated by Q. v.. p., quod erat faeiendum, "which was to be done"). 7. The discussion of special or limiting cases, if such cases oceiir in the given problem, la the figures drawn in eonnection with probleins, tbe (lircii U)ies »fe drawn as hearij lines, the lines required fts light liiu-s, and the oasiliary lines ea dotted lines. y Google .ANE r.EO.MF/J'RY PROPOsri'io>; XXIII. Problem 273, From a given point wilhout a give)i line to draw a perpendicular to the line. Given the line AB and the point 7* outsiilo AB. To construct a pRi-petidicular from thi^ point P to the line AB. Construction. With P as a center and with any oonve- iiient i-iidins, describe an arc intersecting A B in two points aa at C and D. Rost, 3, From Cand J> as centers and with convenient equal radii, greater than J CT>, describe arcs intersecting at Q. Poet. 3. [Assertion]. Then PE- is the 1. required. Proof. P is equidistant from the points C and T>. Constr. Also Q is equidistant from the points C and B. Constr, .-. PR 1 CD. Art. 113. (two joints, eadli eqiiklistant from the exlremifies of a thie, determine the X Mseetor of the line) . Q. E. F. y Google CONSTRUCTION PROBLEMS Proposition XXIV. Problem given line to erect a per- 274. At a given point in pendicular to that line. Kg. 1 Fig. ,3 . Given the point P in the line A B To construct a perpendicular to the line A B at the point P. Method I. Construction. From V (Fig. 1) as a center with a convenient radius, describe an arc cutting off the equal segments PC and PB on the line AB. Post. 3. From C ond I> as centers and with equal radii, greater than PI), describe ares intersecting at li. I'oai. ?.. Draw PR. l-oBt.l. [Assertion], Then PR is the _L required. Proof. Let the pupil supply the proof. Method II. Construction. Take any point (Fig. 2) without the line AB, and with OP as a radius, describe a circuniferenee intersecting the line AJi a.t C. Post. 3. Draw CO, and produce CO to meet the cireumference at R. Draw KP. Posta. I and 2. [Assertion] , Then RP is the ± require<i . Proof. Let the pupil supply the proof. Discussion. When is Method II preferable 1^ y Google BOOK II. PI,AXR GF.OWK'r: Proposition XXV. I'koblem 275. To biseci a given liHfi. Given the line AB. To bisect line AB. Construction. With A and B as centers and with equal radii, greater than i AB, describe ares intersecting in € and t). Post. 3. Draw the line CD intersecting AB in F. Post. I. Then AB is bisected at the point F, Proof. Let t]ie pupil supply the proof. PkOPOSITION XXVI. PltOBLEJl 276. 2'o hisfct a given arc. Given tlie arc AB. To bisect arc AB. y Google CONSTRUCTION PROBLEMS 151 Construction. With A and B as centers, and with con- venient equal radii, describe arcs intersecting at G and Z>. Post. 3. Draw the line CT> intersecting the arc AB at F. Post. 1. Then are AB is bisected at the point F. Proof. Draw the chord AB. Then CD X chord AB at its middle point. (Why?) .-. CD bisects the chord AB, Art. 223. {the X hiseulor of a chord jiimscs Oirovgh the <:':nUr and Inserts tlie arcs siihteiKlal by the chord) . Q. E. F. Proposition XXVII. Probleji 277. To Used <t oitrii aii'jh. Given angle A OB. To bisect angle A OB. Construction. With as a center and with aiij' conve- nient radiu.H, describe an arc intereeeting OA at C and OB at J). Post. 3. With C and I> as centers and with convenient equal radii, describe arcs intersecting at F. Post. 3. Draw OF. Pest. 1, Then lAOB is bisected by line OF. Proof. Let the pnpil suppjv the iit'ooE. )j.a.jf, fii- Construct ail I of iO°, y Google 15a i^ook ii. i'laxe geometky Proposition XSVIII. Problejh 278. At a ginen point in a glvrn .^Imi'jhl Vn-' fo c struct an mujlr equal to a given amjlc. Given thp Z A, iiod the point P iu the line liC. To construct at the point P an angle equal to A A, and having PCfoi- one of its sides. Conatnictiou, With .4, as a center ami with any eonve- nient radius, describe an arc meeting the sides o£ /.A at D and B. Post. 3. Draw the chord T^E. Post. i. With P as a center and with a radius equal to AT>. describe an arc cutting FC at F. Po&t. .t. From P as a center and with a radius equal to the chord DP, describe an arc intersecting the arc HF at G. Post. 3. Draw- FG. Post. i. Then Z GFF is the angle reqnired. Proof. Let the pupil draw the chord FG and complete the proof. Q. E. F. Ex. 1. On B given line oonBtmet a square. El, 2. Construct an angle of 30°. Kx, 3. H«nce eonBtruet an. angle of 15". y Google CONSTRUCTION PROBLEMS iM Pboposition XXIX. Peoblem 279. TliTough a given point without a given straight line to draw a line parallel to a given line. Given any point P without the line A li. To construct a line through P II -4 7i. Construction, Through P draw any convenient line CT) meeting AB in C. Poat. i. At P in the line CD construct / JJPF equal to Z PCS. Art. 278. Then EF is the line required. Proof. Let the pupil snppiy Ihe proof. Ex. 1. Tlirough n given point dvaw a lino paraliel to a given line by conBti'utiting a parallelogram of which the given point is one Ea. 2. On a p;iven line as a diameter, oonKtniot a eirele. Ex. 3. Oonatruet an are of 00" having a radius of 1 In. Ei. 4, On the 9gnrB, p. 148, if the point Q be constructed below the Una JB, will ihe perpendicular required be likely t« be more accurately, or less accurately constructed ? Ex. 5. From a given point on a piven eireumference, how many equal chords can be drawn ? Ex. 6. Throuf;h a p;iveQ point within a given cirenniferenee, how many eqnal chords can bo drawn? Ex. 7. Froru a given point esternal to a given oircle, how many equal secants can be drawn ! y Google r,ANK GEOMETllY PiutrosiTiON XXX. Ph( 280. Tn divide, n tjircn s/migM lim nHiiiber of cqind inni.-<. Given tiie line A B. To divide AB into a required unuiber (as Uiree) eqoal parts. Construction. From ^1 draw the line AT iiutkiiig a con- venient angle with AB. Post. i. Take AC any line of convenient length and apply it to AP a number of times equal to the number of parts into ■which AB is to he divided. Post. 2. From E, the end of the measure when last applied to AP, draw EB. Post. 1. Through the other points of division on AP, viz., C and Ji, draw lines ll EB and meeting AB at F and G. Art, 270. Then AB is divided into the required number of parts at F and 6. Proof. AG = CD = DE. Constr. .-. AF=FG ^ GB, Art. 176. (if three or more parallels intercept cq^ial parts on one transucrsal, Ihey intercept e<pial partu on evenj traimversal). Ex. 1 . CoDstruet a right triangle whose legs are 1 in. and li in 'Ef. %. Congtriiet a J ec tangle whose base 13 1 in. an<3 altitude 1 i: y Google COKSTEDCTION PROBLEMS IDa Proposition XXXI. Problem 281. To construct a triangle, given two sides and the included angle. Given m and n two sides of a ti'iuugle aiul P the augle ineiiided by them. To construct the triangle. Construction. At the point A in the line Afl coustvuet I A equal to the given /.P. Art. 2TS. On the line AJi lay off AO equal to wi. Post, J. On the line AF lay olF .) I) equal to *i. Post, ••. Draw J)C. Post, 1. The A ^I>C' is the triangle required. 1). E. F. Ex, I. Constniat a triangle in ivhk'h two of the sides are 1 iu. snil Itin,, nod the included angle is 13:)". Ex. 2. Construct an isoaeoles tiiun^lu, in wiiiHli Ihe liase shall bii H in, and the altitude 2 in, Ex, 3. Constniet llie complement of a given acute angle, Ex. 4i, Constniet the supplement of a given angle. Ex. 5. How is tiio figure on p, 154 i^onstrueteJ wilh Ihe fewest adjustments of tlie eompasses ? Ex.. 6. DniK a. line (segment) iiud mark olY lliree-iifllis of it y Google 156 BOOK n'-. PLANE GKOMETRV Pboposition XSXII. Problem 282. To consirud a triangle, givni tm angles and ihj included side. Given the A P and Q and the inr^luded side m. To construct the triangle. Consttuction. Take any line AI) anil on it mark AB equal to m. Post. 2. At A construct an angle equal to / P. Art, 278. At B construct an angle equal to Z Q. Avt, 278, Produce the sides of the A A and B to meet at €. Poflt. 3. Then A ACB is the triangle required, Q, E, F. Discussion. Is it possible to construct the triangle if the sum of the given angles is two right angles ? Why ? Is it possible if this sum is greater than two right angles ? Why? n wliiai two of the angles are 30° anii 9 II iii, Ex. 2. Construct the complement of half a given angle. Ex.3, Coiistruot anaDgleo£120°; of I50°i o£ 135°. Ex.4. Trisect a given risht angle. y Google construction pkoblesis Proposition XXXHI. Problem ;S. To construct a triangle, given the three sides. Given m, n tmdp, the throe sides of ;i triangle. To cottstruct the triiingle. Construction. Take the line AB equal to m. Post. 2. With A as a center and with a radius equal to n describe an are, and with B as a center and with a radius equal to p describe another arc. Post. 3. Let the two arcs intersect at the point C. Draw CA and GB. Post, \. Then A ABC is the triangle required. Discu^ion. Is it possible to construct the triangle if one of the sides is greater than the sum of the other two sides? Why? What bind of a figure is obtained if one aide equals the Bum of the other two sides 1 Ex. 1. Cotisti'uet ac aiiijle of 22-i°. Ex. 2. Divide a given cireumlereiiee into four qundranta. Ex. 3. Construct the figure on page 82, using tin the tbcee altitudes aa & teat of the accuracy ot tbe woi y Google 158 BOOK II. PLANE GKOMETltV PiiOPOSiTioN XXXIV. Problicm 284. To construiit <i irkuujh:, given two sidr-f and an angle opposite one of them. Given »i and n two sides of a A and Z 1' opposite n. To conatmct the triangle. Construction. Several cases occur, according to the rela- tive size of the given sides and the size of the given angle. Case T, Wlim n > m {<uid I V U acute) . At the point A construct Z EA1> equal to Z P. Art. 27s. On AE take AB equal to ?h . Post. 2. With £ as a center and with a radius equal to h, describe an arc intersecting AD at C and C. Draw BO and BC. Post. 3. Post, 1. Two A, ABC and ABC, arc obtained, containing the sides m and n; but only one of them, A ABO, contains the ZP. .'. A ABO is the triangle required. Case 11. When n = m (and /.Pis acute). Make the same construction as in the preceding case. The arc drawn intersects the line AD in the points A and 0. Hence the isosceles A ABC is the triangle required, y Google CONSTKUCTION PE0BLEM8 Case III. When n < m {and IP is acute). Make the eonstruetiou in the same way as in Case I. Two ;&, ABC and ABC, are obtained, each of whieli contains the sides m and n and an angle equal to / P oppo- site the side n . :. A ABC and ABC are the triangles required. DiscussioB. In Case I, if ZP is a right Z , let the pupil eoustruet the figure and show that there are two ^ answer- ing the given conditions. If ZP is an obtuse angle, let him construct the figure and show that there is hut one answer. In Case II, if Z P is right, or obtuse, what results are obtained ? In Case III, if ZP is acute and n = the ± from B to AT), how niaoy answei'S are there? also, if w < this ±, how many ? If / P is right, or obtuse, what result is obtained ? Ex. 1. Constniet a triangle in mhieh two of the sides are 1 iu, aud ii in., and the angle opposite ibe latter side is 45". Ek. 2. Construet a triangle in which two of tiie sides are IJ in. and 1 in., and the angle opposite the latter side la 45". Ex.3. Construct a triangle in which two of the sides are U in. snd i in., and the angle opposite the Utter side is 30°. Ex. 4. Construct the figure of page SO, using the coneurreuee of the throe blseutois as a test of (he atjcuraey of thu woik. y Google HOUK II. rLANE OEOME'J'ltY pKOfOSlTlOX XXXV, PifOllLKM 285. To construct a -ihi rail dwj ram, i/Uru /iro sides and the inclmkd uiitjle. Given m and ii two sidos, arid P Uh' iiidiuli-d /. of a ITJ . To construct the parallelogram. Construction. Take line AB equal to m. Post. 2, At the point A eoiistrnet /.OAB equal to ZP. ah, ?7m. On the aide AG lay off AD equal to ". From D as a center and with a radius equal to ))i, Luui from B as a eenter with a radius equal to n, describe arcs intersecting in E. Post. 3. Draw ED and EB. Post. 1. Thun ABED is the parallelogram required. Proof. Let the pupil supply the proof. Proposition XXXVI. Problem 286. To circumscribe a circle about a given triangle. Given the A AliC. To construct a circumscribed O about ABC y Google COXSTEUCTION PliOIiLEMS 161 Construction. Ei'oct Js DFaud EG at tliu niidpoiiits of the sides AV auil AB, respectively. Art. 274. From 0, the poiiic of intersection of these A , witii a radius equal to OA, describe the O ABC. Post, -i, Art. iST. Then O ABC is the eii-cle required. Proof. Let the pupil supply the proof. Q. E. F. PKorosiTiox XXXVII. Problem 287. To inscribe a cinie in a <jhcii Irianffle. Given the A ABC. To construct an Ui^^crihed circle in the triangle. Construction. Draw the Hue AD bisecting ^EAC, and Ci; bisecting- /lECA. Art. 277. From 0, the iutersection of AT) and CE, draw the liiie OP ± AC. Art, 273. From as a center with a radius OP, describe a O . P^-st. 3. Then circle ia the circle required. Proof. Let the pupil [supply tlic proof. Ex. 2. 1 the tLree \ y Google 162 book 11. plane geometry Peopositiox XXSVIII. Problem 288. Throiujh a liiinii jicint on the circumfarence to draio a tanijmt to a cirvk. Given any pniot P on the (jircumference of the circle 0. To construct a tangent to tlie cirele at the point /'. Construction. Draw the radius OP. At the point P construct a line AB X OP. Art. 374. Then AB is tlie tangent required. Proof. Let the pupil supply the proof. 289. An escribed circle is a circle tiuigpiit to one side of a triangle and to the other two sides produced. Thus the cirele ia an escribed circle of the A ABC. A center of an escribed circle, as 0, is called an ex- center Df tho triangle. Es. 1. Draw a triaagle nnd all of its cscribtJ cirtlRs. Ex. 2. Conatnict the figure on page 83, using tlie concurrence of the three medians as a Itst of the accuracy of the work. y Google COXSTKUCTION PSOELEMS lod Proposition XXXTX. Phoblem 290. From a giw.n poini tvithout u circle to drain a iau' gent to the circle. Given F anj- point without the circk 0. To construct a line through F taiigeot to the circle 0. Construction. Draw the I'me PO. Post, : Bisect the line FO at ^f. Art. 37i From M as a center with Jff nri n cumfurcQcc iuterscctiiig the giveu and B. ;, tkseribe a cir- iiftreiiee at A Post. 3. Draw FA and FB. Then PJ ami PIS are tlie tangents required. Proof. Z PAG is iLi>eribea in a semicircle. ■■. IPAO is a ri^rht angle. .-. PA is tangent to tJie circle 0. In like manner PB is tangent to the circle 0. (Why!) (Why?) yGoosle 164 BOOK II, I'LAXE r.Eo:in:TriV Proposition' XL. I'lionLKM 291. Upon a given draUjM liiif to tlr^cribe a segmeni irhich shall conlitin a gh-eii- iiifcrihed angle. Given the ritraight line ^1 n and the Z P. To construct on the line ,17>' a segment of a circle such that any angle iuseribed hi tlie set^uieiit shall equal ^P. Construction. At the point B in the line AB constract ZABO equal to ZP. Art. 273. Constract DE the ± bisectoi- of line AH. Art. 274. From as a centev with OB as a radius, describe the circle AMB. Post. 3. Then AMB is the required segment. Proof, Let AQB be any Z inscribed in the segment AQB. Then /.AQB is measured by h are AFB. Ai-t. 258. But BG is tangent to the circle A2IB. (Why?) .". I ABC is measured by \ arc AFB. fWhy?) .-. /Q= lABG, or IP. (Why?) .*. any Z inscribed in segment A2IB ~ ZP. Q. E. F. . y Google constkuctiox problems Pkopositios XLI. Theorem. 292. To find the common unit of measure of ti surable straight lines, ami hence the ratio of the lines. Given two lines AB luui C7>, of wliieli CT> is tlie shorter. To construct ii common unit of muaaiiro of J U luid CD, and iieiJCR obtain tlie rutb oJ: AB rmd CD. Construction. Apply CT> to AB as many times as possible; Post. 2, Say twice, with a remainder KB. Then apply KI^ to CD as many times as possible ; Poet, 2. Say three times, with a remainder LD. Apply Lll to KB as many times as possible; Post. 2. Say three times, with a remainder I'B. Apply PB to XD as many times as possible; Post. 3. Say it is contaiQed in LD exactly twice. Then FU is the common unit of meiiiuiu of -I /.' :i\v\ I 'H. Proof. LD=2PB. KD=^KP + PB^3LD + PB=1PB. Axs. G, 8. GD= CL + L7)^3KB + i/)=23P7i. ^ B = .4 71 + /r /J = 2 CD + KB = r,3 I'B . A n _ r.r{ 7*/; „ :y.\ Hence -^ - ijj-^. = ^3" yGoosle SXEfiCISES IN CONSTRUCTION PROBLEMS 293. Analysis of problems. The niothcxl of analysis (see Art. 196) is of especial value in tlii; solution of con- struction probleids. In general, to investigate the soiution of a problem by tliis method; Draw a fi<juri; in irJilch the vqi'irfil coiiKli-wtion is ansumed as made ; Draw auxiliary lines, ifnecflssriry ; Observe the rvlaiioHS hdwceii the purtg of this fujiire, in order to discover a known relaiion on ichiclt the required construction depends; Having discovered the required relation, construct another figure hy the direct «se of this relation. Es. Through a, given point within a oii'ole draw a chord which shall be bisected by the given point. Analysis. Let J be the given point witiiinthi? given circle 0, and let FQ be a chord bisected at the point A. A bisected chord suggests a line OA joining tbe point of bisection with the center 0, and that (Art. 224) OA ± PQ. Synthesis, or Dihrct Koluteon. Taking miotlipv IlKure contain- ing the data of the problem, connect the point A with the center of theeircle by the line OA. Through^ dniW a line 1 OA [Art,27+i, nnd meeting the circumfer- ence at tlie points I' and Q. I'Q is tho thurd rL-tinited. EXERCISES. CROUP SS CONSTRUCTION OF STRAIC.nT LINES. Ex. 1. Draw a line parallel to a given line, and tangent to a given [Sua. Suppose the req^uired line drawn; then the radius to the point of tangeney, if produced, is X given line, etc.] Ex. 2. llraw a Hub porpendicular to a given line, and tangent to & given circle. y Google EXERCISES. PEOBLEIIS 167 Ei, 3. Prom two points In the eireumturenee of a eirele, draw two equal and parallel chords. Ei. 4. Through n giren point draw a line which ehall make a given angle with a given line. Ei. 5. Through a giveu point draw n line which shall make equal angles with the sides of a given angle. [Sua. The bisector of the s-Lven I will be ± the required Una, etc.] Ejt. 6. Through a given point between two given pai'allel lines, draw a line of given length with its extremities in the two parallel lines. /'^^'""\ Ex. 7, Through a given point J within a eirele, IA/ "•, \ draw a chord equal to a given line. V \ \ \ Ex. 8. From a given point in the oircumferenco V j_.^ <- 'J ot aoircle, draw a chord at a given distance from ^v.-^^!,-^ the center. Ex. 9. Througli a given point on the circumference of a circle, draw a chord which shall be bisected by another given chord. [SuG. Draw the radius to the given point and on it as a diameter describe a circle, etc. When is the solution impossible ?] 294. Construction of points and of loci. In construct- ing a point to meet certain glvec conditions, it is often helpful to c<nisirw:i the locus of a point iinsu-erhig one of the given conditions and observe in ivhat point or points it meets a given line, or meets anotJier locus ansirei-ing another given condition. EXERCISES. CROUP 13 COXSTBUCTIOX OP POIVl':: AXD LOCI Ex.1. Find a point Pin a given line Jii ^ p equidistant from two given points C and 11 '\^ [Suo. Construct the locus of all points ^ equidistant from C and D and observe where A~=^ \;__„^-_=^ it intersects the given line AB.] ' Ex. 2. Find a point 1' in a given circumfuicuce, which is equi- distant from two givi'n iioiritB, (.' and D. y Google 1 givfu line, which i> Ex. 4. Find a point d, from a giren point. Ex. 5. Find n pnint whioli is at tl given ilistanee, a, from a givfn point A, riiiii at another distance, h, from anothiT givon i)oiiit B. Discuss llio limitations of tliis problem. Ex. 6. Find a point equidititant froia two givnn pointi^, mid at a given distance from a given straight line. [Srr.. Draw the locus of all points equidistant fi'ora the two given points, and also the locus of all poluta at the given distance from tho given straight line, etc.] Ex. 7. Find a point equidistant from two given points, and at a given distance from another given point. Ex. 8. Find a point equidistant from two jriven points, and also from two given intersecting lines. On the other hand, the determiaatioa of certain loci is equivalent to the conslrnciloii of all points which snti.tfy one or more given conditions. Ex. 9. Find the loeua of the center of a aivole, which touches a given line at a given point. [Sro. Construct a number of circles touching the given lino at the given point and observe the relation of their centers.] Ex. 10. Find the locus of the center of a ciraumterenoe with t given radius, r, which passes through a given fixed point. Ex, II. Find the locus of the center of a circle, touching two giver intersecting lines, Ex.12, Find tiie iocusof the c parallel linos, Ex. 13, Find the locus of the ( which toucliBS a given straight line Ex. 14, Find the locus of tiic t which touchcii a given circle, ir of a circle of given radin: ■c of a eirale of given radiu y Google EXERCISER. EROBLEJIS EXERCISES. CROUP 24 COXSTRIXTIOX OF KECTII.INEAK riGL^RE Ex. 1. An eijuilateral triangle, given the pecimeter. Ex. 2. An eqiiilaloral triangle, given the Rltituda. Ex. 3. An isosceles triangle, given tho bnse ami .iltiti Ex. 4. All isoseeles triangle, given the ba.=o mul nri i ?cles ri:mj;'l., Ex. 6. A right tviangle, given n. leg and iho acute angle adjacent. Ek. 7. A right triangle, given li Ic},' and the acute angle opposite. Ei. 8. A right triangle, given tlio hj'potenuse and an acute angle. Ex. 9. A right triangle, given the hypotenuse and a leg. given the altitude and the sides including the Ex. 10, A vertical angle. [SUQ. Through tlie foot of the altitude d of indefinite length, etc.] p 1 altitude and Ex, 11, igle, g iiJes and the ilUtudf? ipon e of Ex. 12. A square, given the diagonal. Ex. 13, .4. rliomljHs, given the two diagonals. Ex. 14. A I'homljus, given one angle and one diagonal. Ex. 15. A parallelogram, given two adjaoent sides and an altitude. Ex. 16. A parallL'lograr .1, giv. ;n a side, the aUltude upon that side and an angle. Ex, 17. A paralUdit-rt im, yi ven the diagonals and an angle included by tlK^m, Ex, 18, A quadrilalerul , b'ive I, th« sides and one angle. y Google 170 BOOK n. PLANE GEOMETRY 295. Use of auxiliary lines ia constructing rectilinear figures. In constructing polygons, auxiliary lines are fre- (iuentlj' of service. Tinis it is often of especial value to c<mstrtici, first, eitlier the inscribed or the eircumseribed circle, and afterward tJie required triangle or quadrilateral. Construction. Drawaeirolo with radius equal to r. Draw a tangent at any point A. On this tangent mark oft All and AC eiicli equal i b. From B and draw tangents iJl'' and CF to tlia circle. TUen BCF is t!io required trianglo. EXERCISES. CROU^ SB CONSTHUCTIOXS. AUXILIARY LINES Coiistruet Ex. 1. An isosceles triangle, givon the base and the radius of the cireurasoribed eiraie. Ex. 3. A right triangle, given (he radius of the iiircumscribed eirele and an acute angle. Ex. 4. A right triangle, given the radius of tlje Inscribed oirele and an acute angle. [SUQ. Draw the inscribed circle and at its center constrnet au anele equal to the supplement of the given angle.} [. 6. A triangle, given the base, flio altitude and tho vertex -ingle. a construct n segment which sliall contain [Boo. On the given the given vertes angle. See Art. 291. J Ejc. 6- A triangle, given the baso, the mL'dian to the base, and the vertex angle. Ex. 7. A triangle, given one of the inscribed circle. tSue Kx. angle, nd the radius y Google EXERCISES. PROELEMH 17X II adjacent angle, and the The use of mtxiliary straight lines may be illustrated as follows : Ex, 9. Constmct a triangl6, ei^'^i the perimeter and two angles. Analysis, Suppose the required triangle JBC already eonstriioted and let AAISC and JCifbe the given angles. Produce £C to D and E, ^ making DB = AB and CE =^ AC. -■-'/'^V- Tban DE= given perimeter. Also --'"'/ \ '""""- Similarly IACB=21E. Hence D B C £ CoKSTBUCTiON, Take D£tlie given perimeter; at 75 ctingtruot an angle = X of 0"6 given angle! "t E eouatriict an angle = % ot the other given angle. Produce the sides of these angles to meet e,t A. Construct ADAB = ID and ICAE = IE. Then A ABC is tbo required triangle, ete. Construct Ex. 10. An iaoscelea triangle, given the perimeter Kud the altitude. [Sue, Bisect the perimeter and eonstruet the altitude ± to it at itsmidpoint-1 Ex. 11. An isosceles triangle, given the perimuter aud the vertoi angle. [SUQ, If the vertL'S Z is known, tiie base i may be obtained.] Ex, 12. Arighttriangle,eiyonnnaciitH ^ angle and the sum of the legs. •' t"""--^ [SOG. Given AB the sum oC the legs, ,-■'' ^''~~--^.^^^^ eonstruet iiA=ii)°, etc.] _^ j) - ■- ^ Ex. 13, A riglit triangle, given an ueuto angle and the difference ot the legs. Ex. 14. A right triangle, given the hypotenuse and the sum of the legs. Ex. 15. A riiilit triangle, given au a«utu angle iind Ihs auoi o£ the liyjotenuse and unu kjj. y Google 172 liOOK II. rL.VNR GEOJIT-.TRY Ex. 16. A triangle, given an s.ng\-:, a siile nM thn sum of ti Other tTTo sides. Ex. 17. A ti'inngle, given an angle A, the sum o£ tlie sides j and BC and tlie altitude upon AB. 296. Reduction of problems. Iti many caao.s a prolxhn may be solved hy reducing the proilcm to a problem alreaii soloed. (Tliis is a special kind of analysis.) Kx. Construct a parallelogram, giycn the diagonals and one side. Analysis: Suppose the £ZJ ABCF to bo Ihfl required /H? alread constructed. Let AF be the given side, If the diagonals ni given, halt each diagonal ia given (Art, 161). Hence in the A AOF the three j? sides are given. Heneo the recinired / "^-.^ _,-■ problem reduces to the problem of con- / -''n'- structing a triangle whose three sides •..--'''' .are given (Art. HS3), Henee Construction. Let the pupil supply tho di EXERCISES. CROUP 26 Rl'IDrCTION OF CONSTRUCTION PROBLEJIS Construct Ex. 1. A right triangle, given the aitituily upon tliy iiypotenusE and ttie median upon the sniue. Ex. 2. A rectangle, given the perimeter and a diagonal (see Ex. 3. A rectangle, given the perimeter and an angle made by the diagonals. Ex. 4, A triangle, given the Ihreo angles and tlie railius of th» circumscribed circle. [SuG. The sides of the A are the chords of 11 segments of the O eontaiuing the given i .] Ex. 5, A triangle, given two aides and the median / / to the third side. j/' y Google EXEKCISES. PROBLEMS li6 Ex. 6. Atrianftle, Riven the thi'ee medians. [Suo. Beauee this to the pveeeding Ex. See Art. 187.1 Ex, 7. An isosceles trapezoid, giveu the buses and ati angle. Bs. 8. An. iaoaceies trapezoid, giveu the biiaes and a diagonal. Es. 9. A trapezoid, given tlie four sides. Ex. 10. A trapezoid, given the bases and tiio two diagonals. [Src. Reduoo to Art. 2S3 by producing, tlie lower base a distance equal to the upper base, etc.] 297. Construction, of circles. The consfcraetion of a required circle is freciuently a good illustration of tlie preceding method of reducing one eonsfcruction problem to another. For the construction of a circle frequently re- duces to tlie prohlem of fiitdhuj a poitit {the center of the circle) which answers giren eonditioiis. (See Art. 294.) Ex. Construct a circle which shall touch two given intersecting lines and have ita center in another given lino. This problem is equivalent to the problem of finding a point which shall be in a given line and be equidistant from two other given lines. (See Ex. 3, p. lOS.) In soma cases, however, the eonstrno- tion of a requirud t^irolii iiiiift be made by an independent method. EXERCISES. CROUP 17 COSSTEUCTIOX OP CIRCLES Construct a circle with given , radius, r, Ex. 1. -Which passes thr. line. ough a given pfiint and tf Kiches a given Ex. 2. Which has its ce' uler in a given line uud t, luehes another given line. Ex. 3. Which passes thri )ugh two given points. CocHtruet ft cirelo Ex. 4. Which touches tw •0 givou parallel lines and passes through a given point. y Google Ex. 5. "Wiiii.-li parses llinin^'li two gi^-Lii on a given line. Ex, 6. Whiuh toucho3 three given lines, two of which are paraHcl. Ei. 7. Which passos through a givon point A and touches a giren lice liC at a given point J>'. [SOG, Draw An tmd at li coiiatriLfl ;i 1 to fiC] Ex. 8. TV'hicli toiithes a givenlins and al: touches s, given circle u / \ a given point ^. ..o Ex. 9. Whieh touchefi .i given line AJl t a given point A and touches a given EXERCISES. CROUP 28 PROBLEMS SOLVED BY VARIOUS METHODS Kx. 1. Through a given point draw a line which shall cut two given intersecting lines so as to form an isosneles triangle. Ex. 2. Construct an isosceles ttiimgle, given the altitude and one leg. Ex. 3. In a given circumference find a point equidistant from two given intersecting lines. Ex. 4. Draw a circle whicli shall toui^h two given inlerseoting lines, one of thein at a given point. Ex, 5. Draw a lino which shall lie terminateJ by the sides of a given angle, shall equal a given line, and be parallel to another given line. 1 ndjac mgle, Ex. 7. Find a point in a, given circumference at a given distanea from Si given point. y Google MISCELLANEOUS EXERCISES. PROBLEMS 175 Ex. S. Construct a parallelogcam, given a siJo, an angle, and a diagoniil Ex. 9. Through a given point within an angle, draw a straight line terminated t>y the sides of the angle and bisected by the given point, [SuG. Draw a line from the verte-i: of the angle to the given point and produce it its own length through the point.] Ex. 10. Construct a triangle, piven the vertt'i angle and the segments of fhe base made by the altitude. [SuQ. Use Art. 291,] n ciidiiis which shall touch a Ex. 11. Construet an isoscclf ertcxand the base. Ex. 12. Drawacircli 5 with si irele at a given point. Ex 13. Construct a right tri; Iti tilde upon the hypoto: QUSO. ■, given the hypotenuse and the Ex. 14. Construct a triangle, given the base and the altitudea npon the other two sides. [SuQ. Construct a semieircle on the given base as a diameter.] Ex. 15. Find a point in one side of a triangle equidistant from the other two sides. nd the angles at Ex, 17. Construct a rhombus, given an angle and a diiigonal. Ex, 20. In a given circle draw a chord equal to parallel to another given line. [SuG. Find Ibe distance of the given chord fro: couattiieting a right triangle of which the hypotouuse given,] y Google 1(G 1iOOK II, PI.AXE GEOMETKY Ex. 21, Consfrupt a ti'in!if;Io, giv'ii nil aiL.ijii;, thu bisei'tor of that angle, and tbe aititude from iiuotlitir vertox. Es. 22. Find the loeus oE the points (jE ciontact oE tangL'tils dfawa from a given point 1o ii hltIl's of circlea kaviii!!; a given i^tuter. [Sm, Uao ArU. il'JD mid 2til.] Ei. 23. Given a line ,17; ami Uvo points. -G C and D, on the same side of JJl. Find a point P in ^J! such that IAPC= IJU'D. [Suo. Draw a 1 from C to J7! ajid pro- j^ +^ ^ dueeit its own length, ecc] Ex. 24. Given a line -17i and two points r and }> on the samo side of AB; find a point P in J7( suth that CP + I'D sliall ho a minimum. Ex. 25. Draw a tangent to two given eireh Ex. 26. Dra' tangent to two given circles. y Google Book III PROPORTION. SIMILAR POLYGONS THEORY OF PROPORTION 298. Ratio has been defined, and its use briefly indi- cated in Arts. 245, 246. 299. A proportion is an eKpressioTi of the eriiiality of two or more equal ratios. As, This reads, "the ratio of d to J equals the ratio of c to d," or, "a is to i> as c is to d." 800. The terms of a proportion are the four quantities used in the proportion. In a proportion the antecedents are tlie first and third terms; the consequents are tlie second and fourth terms; the extremes are the first and last terms; the means are the second and tJiird terras. A fourth proportioaal is the last term of a proportion (provided the means are not equal). Thus, in « ; fe = c : d, dis a. fourth proportional. 301. A continued proportion is a proportion in which each consequent and the next antecedent are the same. Thus, a : & = & :c = c : d = d : ^ is a continued proportion. A mean proportional is the middle term in a continued proportion containing but two ratios. h (IT") yGoosle 1(8 ISOOK III. PLANE GEOMETliY A third proportional is the last term In .i coiitiimod pro- portion containing but two ratios. Thus, m n : h^h : c, b i& a. mean pi'oportioual, and c is a third proportional. Phoi'OSitios I , Theorem 302. In any proporiion, the prndiict of the extremes is equal to tlie pvodwl of tlie wniiis. Given tlie proportion a : b = r : d. To prove ad = be. Proof. f^ = 2- "^i"- Multiply each memher by hd. ad — he. Ax. i. Proposition II. Theorem 803. TJie mean proportional iHween two quantUie; equal to ike square root of I heir product. Given tlie jtroportion a :l) = h : To prove b = y( Proof. a:l>^h : :. h" = ac, (in any projfortion, the i>rodiict of th- .-. b=v;^ Hyp. Act. 303. equals the product of the Q. E. B. Ex, 1. Find the fourtli proportional to 2, 3 and C ; also to 3, i, ^. Ex. 2. Find tlie mean proportional between 3 and 6. Ex. S, Find the third proportioual to 3 and 5. y Google THEORY OF PROPORTION 179 Proposition III. Theoresi 304. If the product of two quantities is equal to the prod' net of two other quantities, one pair may be made the ex- tremes and the other pair the means of a proportion. Given ad = hG. To prove a , b^c: (I. Proof. ml- Ik. Divide eacli memlici l.y M. Then b' d Or o :S-C ; <!. 306. Cor. 1. If the (uitecedeuts of a proportion are equal, the consequents are equal. Thus, if a:x = a\y, then x = y. Let the pnpil supply the proof. 306. Cor. 2. If three terms of one proportion are equal to the correspondinrj three terms of another proportion, the fourth terms of the two proportions are equal. Thus, if II : h = c : x, and a : li = c -. y, then x = y. Let the pupil supply the proof. Ex. 1. 'Write <ih = i>q as a proportion in as many different ways as Ei 2. Write a:(j:+l) = (i na a proiwtiou ; ulao j;- = 15. y Google Proposition IV. Tueokem 807. If four quantities are in proportion, they are in proportion 6j^ alternation ; that is, the firnt term in to the third as the second is to tlie fourth. Given the proportion a •.b = c -. d. To prove a . c=h -. d. Proof. a : h^c : d. Hyp. .-. (id = bc, Art. 302. (in anu pronortimi, ike product of the extrnauis equals the iiroduct of Writing a and d as the extremes, and c and h as the means of a proportion, a : C = l) : d. Art. 304. (*/ the product iif tieo qnaiitities is eqiiiil io ih^ prodni-t of iiro nlher Proposition T . Theorbm 308. If four quantities are i» proportion, they are in proporHow 6j/ inversion ; that is, the second term is to (he first «s the fourth is to the third. Given the proportion a ■.h = c -, d. To prove Ji : a^—d -. c. Proof. a : h — C: d. Hyp, .-. ad^bc. (Why?) Writing 6 and c as the extremes, and a and d as the means, I : a^d : e. (Whylj Q. E, ». Ex, TranB£ormi:a = i):o bo that j: shall be the last term. y Google TlfROliY OF PROPOKTION 181 Peoposition VL Theorem 309. If four quanfitips are in proportion, they are in proportion by compositioa ; iktif ix, the. snm of the first two terms is to the second term ws the sum of the last two terms is to the last-term. Given the proportion a : h — c -. d. To prove a -{- h -. h^c-V d: d. Proof. 7= T ^?P- h a Add I to oach meiuber of the equality. As. 2. h a ^ b d That is a + h -.h^c + d: d. Let tlie pnpil show also that a -r h : n = c-{- d : c. Froporitios- VIT. Theorem 310. If four quindiUcs are in proportion, they are in proportion ?!(/ division ; that is, the difference of the first two is to the second as the difference of the last two is to the last. Given the proportion a -. b = c : d. To prove a — b:l> = c — d:d. Proof. ^ =^- Hyp. yGoosle 18'J HOOK iTt. i=LA>rE nroMi Subtratit 1 £min vM-h mi>,m)Mv of I he • 1.1 d a — h '■ — >! That is fl-?i : i = c — d : (L Q. E. D. Let the pui>il hIkiw also that a^b • a—c — d c. PRorosrnoN VTII. Theorem Sll. If Jovr quaniities are in proportion, they are in proportion hy composition and division; that is, the sum of the first two is to their difference as the sum of the last two is to their difference. Given the propoi'tioii a -. b = c : d. To prove a-\- b -, a — !> = <■ -\- d -. r, — d. Proof. *' ; /' = '■ : 'I. Hyp. ^'-^h_c^d_ By cooi|msifii! d Dividing the coiTespoiiding men equalities, a+b^ c+d a — h e — d' That is a-^ h : a — & = c + d -. Ex. 1. WhHt do«e the proi-iortion 12:^ = 8:2 beeoi tionT also ty divisiout Kl. 2. Whatdoea 2.c — 5:5 = 3^: — 7:7 becomoby t y Google THEOBY OF PROPORTION lod Proposition IX. Theorem S12. In a series of equal ratios, fhe svm of all (he ante- cedents is to the sum of all the consequents as any one ante- cedeni'is to its consequent. Given a -.l^c : d = e if^g : h. To prove a-\- c-\- e-}- g -. J> + d+f+ h = a -. h. Proof. Denote each of the eqnul ratios by r. Then l^r, :. a^br. Similarly c — dr, e=fr, g — hr. Ax. i. Hence a + c + c + q={b -\- <l -i- f + h) r. Ax. 2. Dividing by b i- d +f+ h, ^qrf+y:^ = *" ^ ^ ' ^'^- ^■ That is a + c-<re + g -.h + d + f+h^^aa. Proposition X. Theorem 313. The protiucts of the correspovllut/ /ci-Jiis of two or more proportions arc in proportion . Given a : I'^c : d, e : f^g -. h, ami j : A-=i : m. To prove aej ; hfk^cgl : dhn. Proof. 7= -,. 7= f , T= -■ Hyp. h d f h k. i)i Multiplying together the uorvetipoiiilijig terms of these equalities, Ai. 4. Q. E. B. aej _ cgl t)fk dh7n That is aej : bfk^cgl : dhm. y Google 54 EOOK III. 1>T,ANF. CJEOMETIiV Proposition XL Theorem 314. Lilie 'powers or like roots of ilie terms of a proper- on are in proportion. Given the proportion » -. h^c -. d. To prove o" : V ^ c" ■. i!", and a" : h" = v" -. d". Proof, 7= V Byp, Raiding both loembers ui the h"' power, That is a" -.h" =c" :(!". In like manner ci" -. ?i" = c" : d". Proposition SII. Theorem 815. Equimultiples of two quantities have the same ratio as the quantities themselves. Given the two c|naiitities a and 6. To prove ■ma -. viJi = a -. b. Proof. 7- J- Went. Multiply eaeh term of flie first fraelion by m. ina_ a ■'■ ml>~ h That is ma : mh = a -. h. __^_________ Q. E. B. Ex. 1. Transform ji : q = x : j in all possible ways by the use of the properties of a proportion. Ex. 2. TraQB£ovm 7:3 = 28:12 by eompoaitiou aiiJ diTiBion. Ex.3. Also2a; + 5;2i-5-a^+l:j;' — 1. y Google THEOBY OF PEOPORTION 18.) 816. Note. In the theory of proportion as just p^eaented, the CjuaQtities used are assumed to be ooinmenaurablo, but the saiue theoreniB may also be proved for proportiona wlioae termB ara iucom- menaurable by use o£ the method of limita. For each incommenan- rable ratio may be made the limit of a eorrespondiug eommenaurable ratio; then, by ahowing that the variable commensurable ratios are equal, it may be proved that the limiting incommensurable ratios It is also to be noted, that, in the above theorems, the terms of a ratio must be of ttie same kind of quantity; that is, both be linea, or both besurfacus, etc. Hence, in orderthat aproportion be treated by alternation, for instance, all four of the terms must be of tlio same Ex. 1 . Find the fourth proportional to a, 2<i, 3 ic. Ex. 2. Find the third proportional to rt -;- '< and a — h. Ex. 3. Find the value of x in the proportion, 4 : 5 = a ; 15. Bx. 4. Findft shortmethodof determining whether a given propor- tion is true or cot. Use this method to determine whether the follow- ing proportions are true: (1) 4 r 6 = 3 : 9. (2) 5a : 2o = 15 : 6. Ks. 6. Construct exactly the figure of page 77 with the fewest pos" sible adjuatmenta o£ the compasses. Ex. 7. Draw an obtuse triangle and uaing the eoncuiTeuce of the altitudes as a t Ex. 8. Arrange nine points in a plane greatest number o£ straight liues may pasa i: pass through three points and only three. y Google EOOK III. TLAXR flFllMETP.Y PRorosiTiox Xlll. Tin:oBKM 317. A line panilhl l'< ime, -ihlc of a triangle and meet ing the other two skUs, dlfides these sides proportionally. Given the tviaiiKle ABC ami the Urn- DK \\ base liC and intersectintr the sides AH and AC in ilie points 1) and J-], respectively. To prove BB -. AD^EG : AE. Case I. When T>B mid AD (Fig. 1) are commensxirabje . Proof. Talie iiny cornnion unit ..f niKisiiro of I)B and AD, as AS', and let" it be .'Oiitainu.l iu /)/.' a cerlain lunnWv of times, as n times, and in AP, m \\mv^. DP, Tbroiigli the points of division of DP. and AB drai^ lines II BC. These lines wiil divide EC into i\ . and AE into m parts. all equal. EC_n •'• AE in DB^EC AD AE Art. 17G. (Why ?) y Google PROPORTIONAL LINKW 187 Case II. When DB and AD (Fig. 3} are incommenKurnlile. Let the line AB be divided into any number of equal parts and let one of those parts be applied to DB. It will be contaiued in DB a certain onraber of times, with a line PB, less than the unit of measure, as a remainder. Draw PQ \\ BG. Then DP and AD are commensurable. Constr. AD AE '^^' If now the unit of nieaKure be indefinitely diminished, the line PB, which is less than the unit of measure, will be indefinitely diminished. Hence DP ^ DB, and EQ ^ EC us alimiL Art.'isi, DP - AD EQ ) ^^„.. .. ........ ^ DP ^, . -,-, EQ — ^thevanablej^a But the variable ;. the limit ^-^,= tho limit f|- Avt. ^.w. AD AE d. E, D. 318. Coit. 1. By eoniposition. Ail. ".09. DP + AD : AD^EC -{■ AE : A E. Or AB : AD = AG i AE. In like manner AB -. DB = AC -. EG, or, in general language, if a line parallel to the base cut the sides of a triangle, a side is to a segment of that side as the other side is to ilie corresponding segment of the second side. 319. Cor. 2. Using Fig. 2, m^AL I!^=A^ PTJ^DP QO AQ' ^^^^ EQ AQ " QG E</ Henee. if two lines are cut by a number of parallels, the corresponding segments are proportional. y Google 188 IKKlK III. PLAXE GROMETlii- Proposition XIV. Theohi-.m (Convehse op Vmn'. XlII) 820. If a stffiinht line diviiUn iiro si'h-n of a triangle proportionally, it is iMraUel to the third side. Given the A ABQ and the linu DF intersecting AB and AG BO thut AB : AB^AG : AF. To prove DF]] BG. Proof. Through 7> draw the line BK \\ BG and meeting the side -iC in K. Then AB : AD = AC : AK, Art. sis. (if a line II bam i^v,t lite sides of a a, " aide is to a segment of ilial side aa tii^ other side is to the corresponiling segment oj the second side). But AB : AD-^^AG : AF. Hyp. .*. AF^AK, Art, 306. [if three terms of oni! ^roporthm are cijiinl to the enrrespouding three terms of another prn/wtiiHi, Ihr.fiiiirlli terms of She tiro ' j,roi>orUv)is arcapial). Ilenee the point K falls on F, and the line 1)K coin- cides with the line DF. Art. 6C. But the line DK 11 BC. Constr. ,-. BF !! BG, (for DF coincides Kith DK, ivkicii is || BC). Q- E. D. Ex. 1. In the figure of Prop. XIII, if AD = 6, DB = 4, and ^£=9, find EC. Ex. 2. Also, if AD=12, Dii = S, aad AC^la, find AE aaii EC. y Google EXERCISES. CROUP IB, REVIEW ESERCISKS Make a list o£ the propertii': Ex. 1. One straight lin be related to the line). Ex. 2. Two straight lines that meet, or intei With such angles as may be formed by the given, li Ex. 3. Two or more para,llel lit and angles formed). Es. 4. Eight angles. Ex.5. Complementary angles. Ex. 6. Supplementary angles. Ex, 7. A single triangle (in triangle, aa altitudes, medians, et Ex. 8. A right triangle. Ex, 9. An isosceles triangle. Ex. 10. Two triangles. Ex, 11, Two right triangles, Ex. 12, A rjuadrilatersl. with RHph points as may nnneotion with lines within the Ex. 13. A trapezoid. Ex. 14. An isosceles tvapeioid, Ex. 15. A parallelogram. Ex. 16. A rhombus. Ex. 17. A polygon. Ex. 18. A single related poi' flts). Ex, 19. Arcs of angles) . Ex, 20. Chords 1 Ex. 21. Central a Ex. 22. Tangents Ex. 23. Seeants 1 Ex. 24, Two ciR' niferei (ir rotated lines ai ^-ith related lie y Google 190 HOOK in. PLANE OEOJIETRY SIMILAH POLYGOWS 321. Bef. Similar polygous are polygons having their homolosroiis Liiigliis et^iial and their homologous sides pro- portiooal. Thus, if tliR figures AB€7)E and A'B'C'B'E' are simikj, the angles A, IS, 0, etc. , mnst equal the angles A', B', C, etc., respectively; also ^B : A'B'^BC : B'C'^ GJ) -. OB', etc. Hence it is cOiiMtantly to be borne in iniud that simi- larity in shape or form of rectilinear figures involves two distinct properties: 1, The liojiiologoiis angles are equal. 2. The homologous sides are proportional. It should also be clearly realized that one of these properties may be true of two figures, and not the other. Thus, in the rectangle A and the rhomboid B, the cor- responding sides are proportional but the corrt angles are not equal. !i Also, in the reetangle G and the square J\ the corres- ponding angles are equal but the corresponding sides are not proportional. However, it will be found that, in the case of triangles, it one of the two properties is true, the other must be true also, 322. Def. The ratio of similitude in two similar figures is the ratio of any two homologous sides in those figures. y Google SIMILAR POLYGOKS Pboposition XV. Theohem 323. If two triangles are mutually equiangular, they are similar. Given t!ip & ABO and A'B'C witli lA^ lA', IB = AB', and / 6' = Z t". To prove tlie &^ ABC and A'B'C similar. Proof. Tlic given & are niutuaEly equiangular. Hyp. Hence it only remains to prove that their homologous sides are proportional. Art. 321, Place the A ^.'S'^upon the A ABO, so that I A' shall coincide with its equal, the ZA, and B'C take the posi- tion FH. Then lAFR = IB. Hyp. .-. FHWBC. (Why?) .-. AB : AF^AC : AR. Art. 317. Or AB : A'B' = AG : A'C. Ax. 8. In like manner, by placing the A A'B'C upon the A ABC so that the Z S' shall coincide with its equal, the ZB, it may be proved that AB -. A'B'^BC : B'C. Hence the A J,£Cand A'B'C are similar. Art. 331. 324. Cob. 1. If two triangles have (wo angles of &ne equal to two angles of the other, the triangles are similar; also. If two right triangles have an acute anijle of one equal to an acute angle of the other, the triangles are similar. 325. Cor. 2. If two triangles are each similar to the same triangle, they arc similar to each other. y Google 192 BOOK III. PLANE GEOMiyi'EY Proposition XVI. Tiieohkji 326. Jf iieo triauffles liavc their }w>nolo<jous sides pro- porlional, Ihey ure. similar. Given tlie A ABC and A'B'C, in which AB : A'li'^ AC: A'C'^BC: B'C. To prove the A ABC and A'B'C similar. Proof. The given & have their homologous sides pro- portional. Hyp. Hence it only remains to prove that the A are mutually equiangular. Art. 321. 1. On AB take AF equal to A'B', and on AC take AS equalto A'C. Draw Fff. Then AB -. AF^AC -. AH. Hyp. .-. FH II BC, Art. 320. (if a siraiglit line dickies two sides of a A proportionally, il is || the third side). :. ZAFU = IB, and IAHF= I C. (Why?) ,•, A AFH and ABC are similar, Art, 323. {if tico &. are muttiatly equiangular, they are similar) , 2. .■- AB : AF=BC : FR. Art. 321, Or AB : A'B' = BG : FH. A:^. 8. But AB : A'B' = BC : B'C. Hyp. yGoosle SIMILAK POI-TGONS Hence the A AFH and A'B'G' are ecinai. .-. A A'B'C is similar to A ABO. [for ils equal A AFS is similar to A ABC). Proposition XVII. Theorem 327. If two irianyles have an angle of one equal to an angle of the other, and the inciiidhiy sides proportional, the trianglea are Kimilar. Given the A AHVuud A'B'C, in wiiit^li Z.A^ /.A' and AH : A'B' = Aa: A'V. To prove tlie A ABC and A'B'C similar- Proof. Place the A A'B'C upon t.lie A ABO so that ZA' shall' coincide with its equal, the Z.A, and B'O' take the position FH. Then AB : AP^AG : All. Hyp, Hence FII \\ BO. (Why?) .-. Z AFII = /.B, D.n<i ZAEF= ZC. (Wiy!) .'. A ABC and AFH are similar. Art. asrt. Or A ABC and A'B'C are similar. Ax. 8. (). E. D. miliirA.i'ii'O', y Google Proposition XVIIl 328. If iivo tnangles have tK.. ^ pp.ndiciilar, each to each, the triangles are similar. Given tlip A A'B'C (Fig. 2) with it; Theorkm S'i'les parallel, or per and tliM ■espondir A A'H'ff (Pig. 3) with its sidos -L, nf the A ABC. To prove & ABC and A'B'C similaf. Proof. The A A and A' are either equal or supple- mentary, {foi-llie»i<lc3fm-miig tJ,emn:-e\\ or X). Arts. 130,13?. Similarly, the A B and B', and C and C are either eqnnl or supplementary. Hence one of the three following statements must be true eoneerning the angles of the A: either 1. The A contain three pairs of supplementary A and lA^ lA' = 2vX. A. IB -^ IB' ^2 rt. A, AC + AC' = 2 rt. A ; or 2, The A contain two pairs of supplementary A , as /.4=ZA', ZB + Zit' = 2rt. A, I C + AC = 2n. A : t,r :i. The A contain three pairs of equal A and AA = AA', IB=IB'. AC=AC, Art. 139. (if two A of a A = two A of another A, the third A are tqnai). The first two of these statements are impossible, for the sum of the A of two A cannot exceed four rt. A . Art. 134. Hence the third statement is true, and the A ABC and A'li tj MM liiutuaily cqiiiungular, and therefore Bimihir, y Google Proposition XIX. Theorem 329. If iiro pohjijmin arc xiMilci-, ihcij may he separated into the Sfime number of iriaHglen, similar, each to each , and similarly placed. Given tlifi similar polygons ABODE and A'li'C'JyE', dividml into triangles by tho diagoiiids AC, AD, and A'C, A'jy drawn from the corresponding vertices A and A' . To prove tliat & ABC, ACD, ADE are similar U) tlie ^ A'B'C, A'Cry, A'D'E', respectively. Proof. 1. /.li ^IB'. Ait.:m. Also AJi : A'B'^Tin B'C. Art. r.'Ji. ;. AABV and A'B'C are similar, Art. yy.. U/Uco ii l,i(fc<iii L of one = <m Z of the i'lhe,- >„;l the ii,<-l,idim; 2. Again IBGI) ^ IB'C'D', Art.s-i. (JiwB(0io(70ns i of similar poliigojin) . Also Z BCA = Z i"C".4.'. Art. 321 , {homolo!i,ms A of tlie Similar A AlSt: <uid A'H'C). Snbtracting lACD = lA'C'D'. An, a. But £0 : B'G'=CD : C'Jf, Art. 31!i. {homologous sides of siiiiiltir pnJi/goiis are proportioTial). And TiC: /i'C = ^lC: ^'G', Art. 321. (iMjijr homologous siili^s of the similur & ABCmd A'B'C). Hence jIC : A'C ^ CD -. CD'. Ax. i. :. the ^ A6'/J and A'G'iy are similar. An. 327. 3. Ill like manner it oau be shown that the A .iDE i^nd A'D'E' are similar. q. z. b, y Google BOOK in. PLA-NK lir.i: PROK)tiITIOX XX. THEDliEJl (CONVKKSE or PliU!'. XIX) 330. // two poli/gons are cotiiposi'd of the name, nmnhvi- of triangles, similar, each to each, and similurly placed, the polygons are similar. Givea the two polygons ABCDE and A'B'OB'W, m ■wliieh the A ABC, ACJ>, AD-E are similar, respeetivelj", to the A A'B'C, A'C'JV, A'lyS', and are smiilarly plaued. To prove the polygons ABCJDE and A'B'C'D'E' similar. Proof. /.B = ZB', Alt. 321. {hfiiKj homologous S of similar &.). Also ABCA ^IB'CA'. And lACD^lA'C'D'. Adding PC(iials. ZBOl) = I B'C'iy. In like manner it may be shown that Z Cl>E~ Z (■'!>' ?J\ ZBAE= IB'A'E', etc. Hence the polygons are mutually equiangular. ., AB BG ,, . ., ^ . ^^^'^ .IIP, " "STT^ (homologous sides of mi {Why?) (Why-, (WLy.^ Art. 321, Again - In like n BC B'C'~ CB CI)'' (Why?; A'B' B'C BG_^AC_ CD ^AG ^ B'C A'C" C'ly A'O CD ^BE __^A E CD' D'E' A'E'' Hence the homologous sides of the given polygons an proportional. .-. the polygons ABGDE and A'B'C'D'E' are similar. Art. 321 9. X. B. y Google I'ROl'OR'J'lOSAL LINES 331, Note, It is often eonvsment to write a saries of ( ratios, like those uspd ia the above proof, us follows: AB^^BC _/'AC'\ _ £D.^('£0^_ DE _AR A'B' B'C \A'C'J~ azf yA'D')~ D'E'~ A'E'' a ratio which ia used merely to show the equality of two other r being inclosed in parenthesis. Proposition- XXI. Thkokem 332. In any triangle-, the bisector of an angle divides the opposite side into segments lekich are proportional to the other two sides. Given the A A H(\ with the line A l> bise^^ti[lg the Z BA C, Jitid meeting BC at /', To prove ftC ; !>H^A<! -. AH. Proof. DriiW the line CF\\AD, and moetiiijc ^1 />' pro- dnee,d at F. Then 1)€ : DB^AF : AB. Art, 317. But Zr^lp'. Art. 124. And ls= £p. Art. i:!ii. Also £p'— Lp. Hyp. .-. Lr= Is. (Why?) .-. A .40Fi8 isoseeles, a.m\AC=AP. (Why?) Buhstituting .-1 ('for itsec|uiU, A-F, in the above proportion , DC : DB^AC: AB. An-, s. Q. E. B. Ex. If. ill tlie above figure, AB^U.AC- VI. nud /ir r M, find DC and Di:. [SvG. Let lJC = Jr. /V/f-U — 3, etc.] y Google )S HOOK .III. I'LANi; (;eu.metky 333. A line divided internally Ua Vuiv divideil into two irtK by ii Doint ty.ken betvveen its extremities. Tims, tlie line Ali is divided intpriially at I into tl)e segments PA and PB. To divide a givi'n line interjially in a frU-Pii ratio (as ' 334. A line divided externally is ii line wliose parts ai considered to bo llie segments included between ;t ]ioiiit (j the line produced and the extremities of the gi^■cu line. Thus, the line AB is divided externally at the point. /" into the two segments P'A and P'B. To divide a given line externally id a given ralin (us 2 : T), divide it ratio (bkT' — 2, (ir .) fiarts) uiid ]iiMdiu;e it lill liic pmdutfd pait ciiuals tlie siualler ti.rm of llie ratiu ihiK^ ilie unit line IVmud. 335. A line divided harmonically is a line divided Imih inlprnally aud exUTiuilly iu the same ratio. Thus, if the line AB is divided internally, so that PA .- PB-l : 2, and externally, so that FA -. /"/>'=! ; 2, then PA : FB = PA PB, and the line is divided harmonieally Ex. 1. Divide a givea line barmonicallv in the ratio I ; 3. Rs. 2, Divide a given line liarniaiiiciilly in ilio ratio ^^ : j. y Google rKOPORTIOKAL LINES JLH) Proposition XXII. Theokem 386. In any triangle, the hisector of an exterior angle diriihs rxiernalhj the opposite side prodiiird into segments which arc proportional to the uther two sides. Given the A AUG with its sxtei-Lor Z UAO bisected by le line AD ineetius tbe opposite side produced at D. To prove 1)B: DC^AB: AC. Proof. I ru\ - the line BF II AD and meeting AC at F. Tliyii DB: OC^AF: AC. (Why!) But Z/' = Z/. fWbyr) Aud Z. = Z/.. (Whj!) Also lp'=Zp. .: Zi-= Is. (Why?) (Why!) A ADFia isoseeles, and AB = AF. {Why}) Snbstiti : AB for its e(inal, AF, 11 the al) ve ]>vo- ortiou. DB -. DC^AB : AC. i\\. S. 337. Cob, The lines bisectttxj the iiUn-ior and ixtcrhr anf/lis of a Iriaii'jk at a given vertex, divide- the opiiusite siiJt karmoiiicully. y Google Proposition" XXIII. Theorem 338. The homologous altitudes of two similar ti-ian'jlet have the same ratio as any two homologous sides. Given the similai- A ABC, A'B'C, und IW, B'J)' any two homologous altitudes in these ife. B I) B A BO A a To prove WD' = WA=Wo'^Tr/ Proof. In the rifrht A ABD and A'H'iy, Hij hiiiimhigiius A of tbesiniilfir & JlWanil A'B'<''). :. A ABI) aiul A'B'I)' are similar, Art. -.i-^A, A hace an acuta L of one — an acute I of the other, the & ■• B'jy B'A'' ''^''^' ■'"'■ 1 the similai' A ABC and A'l.i'C, BA BC A C Wa'^Wc'^a^' ^"■^^'■ b i> ^ ba ^bc ^a c _ b'd' b'a' b'c a'c' ^"^ ^' <). E. ». (, AC=18, A'C = 12, and tlD = y Google I-ROPORTIONAL LINES 201 Proposition XXIV. Theorem 339. If three or more lines pass through the same point and intersect two parallel lines, they intercept proportional segments on the parallel lines. Given the transversals 0^ , OB, OC, 01) intei-secttnE the parallel lines Ali and A'D' in the points A, B, C, D and A'. B', 6", !>', respectively, Ali B a (J n To prove a1^'=JF(J'^c1^- Proof. A'D' 11 .1 P. R.rp. Therefore the hiise A. of the A A'li'O, B'C'O, et,e., are equal to corresponding base A of the ^ ABO, BCO, etc. All, nti <n- Art. [2.1. Hence the A A'B'O, B'C'O, etc, are similar to the A ABO, BCO, etc., respectively. Art. 324. iVB' Kii'o) B'V \V'Oj C'l)' ^ '■■ Es. If tbe skies of a polyt,'oii 111* polygon, the side homologous the second polygon. y Google MJ, BOOK III. I'LASE GEOJIKTllV Prop. SXV. Theokem {<Jonvekse of Prop. XXIV) 340. // three or more non-parallel straight Unes inter- ccpi proportioHul si'/jmrnts oh two pnraUds IJiey pass through the name point [thiiC -is, are coiu-iirreiil). psirallel lines AC and A'C f that - Given the transversals AA', BB\ CC intersecting the' AB ^ BC ' A'B' B'O' To prove that the lines AA' , BB' , CC are eoiicurreut. Proof. Pt'odnoe the tines AA' and BB' to meet at some point 0. and let it intersect t )nnv the lir B point r. A'V Al 1 H IW But HeDue A'B' B'P A'B' B'C B'1'=B'C'. Art, ; .'. point P coincides with point C. .: line CC coincides with line CP. Art. :. the line CC, it produced, passes through 0, (for it coincides with the line CP, iBliich passes through 0). :. AA', BB', CC meet iu 0. y Google PEOPOBTIOXAL LlXh;s 203 Proposition XXVI. Theohem 341. The perimeters of two similar pohjijotis have ike same ratio as any two homologous sides. GiveE t Iw two Jiimiliir i">lyt,^uLi.s A IIUDE and A'/i'VITI!', with their perimeters denoted by P and P' Slid with AB aad .I'/i" itny two homologous aides. To prove that F -. r = AB : A'li'. Proof, AH A'H'^BC : li'V^CD : G'I)'^<-t<i. Art. 321. Hence An+IW+Cn+,i-tv.. : A'fr+ B'V'+V'I>'+,etc., ■^Ali : A'B\ Art. 312. (iii a wnra of rt/ml rirfio.'^. Ilin s,:,„ <,/ ,<n liir t,:lri-e<lenl>! !« /" Ihrn^im „J <\ll Ihc eomequcnls as amj m,t ,u,UT.cilcnl is to its cons<:>pieuf). Or P:/" = -t/i ; A' !S' . Q. E, D. 341 (CI. Col;. !i, tiro :<i mile, 'j.ohii/on--^. uiiij two huuwl- iKjoax lini-i air hi rack other as iinij other lii'o hoiiioLoijoiis liutx; and the (xriiwter^ are to each other asainj two homologous lii,(n. Ex. If the porimuUT of a dv.'ii li.'ld is 210 ridf^ ami i. siilr «t ihia Held is ti. !L himioloHdus Ad<- of a similar li.Od iii ^ ; 2. tioil Uie i>crmie- ter of the iouiucl litlil. y Google Proposition XXVll. Theorem 342. In a right trianijle, I. The altitude to the hypotenuse is a mean proportional hetween Die segments of the hypotenuse; II. Each leg is u mean proportional between the hypote- nuse and the segment of the hypotenuse adjacent to the given leg. Given thfi Hglil. A ABO and BF ihe altitude upon the hypoteniiMe A G. To prove !. .IF: HF^HF-. FC. I A<! : An = AH: AF \ AC : BC^ liC: FC. Proof. Th(! Z A is coiriiiinn to the rt. A -4 BF and ,4 BC. .: A ABF and ABC are similar. Art. :i*J4. Similarly Z <7 is common to the rt. A BFC and ABC, and A BFC and ABC are similar. .-. &. ABF" and BFC are similar, Art. 3-25. {if two & are similar to the sam« A, they are similar to each other). Hence, I. In tlie A ABF and BFC, AF : BF-= BF : FC, Art. a21. {homuhfjin'f suies of simitar & an: pruporUonal), y Google PROPORTIONAL LIXES "JUS n. In the simiiar ^ ABC uiid ABF, AC : AB = AB : AP. Art. 321. Also, ill the similar A ABC and BFC, AC:BC=BC: FC. Art. 321, Q. E. B. 343. Cor. The pet-pendicular to the diameter from any point in the eircumfercnce of a circle is poriional ieiween the segmeiiis of the di- ameter; and the chord joining the point to an extremity of the diameter is ,a i proportional heltveen the diameter and the aegmcnt of the diameter adjacent to the chord. 344. Def. Tbe projection of a point upon a line is the foot of the perpeudicular drawn from the point to the line. Thus, if AA' he perpendicular to LM, A' ia tiie projec- tion of the jHiint A on the line LM. 345. The projection of a line upon another given line is that part of the second line which is included between per- pendiculars drawn from the extremities of the first line upon the second line. Thus, the projection of AB on LM isA'B'; of CD on PQ. is CT). Ea. 1. IE the si!|-mentB of tho b.V[ioteimae of a right triangle are 3 and 12, find the ultitiide on tlie hypotenuse ; also find tha legs of the triangle. Ex. 2. If, in the figure of Art. ^43, the ciiainetet is 20 and the longer chord IK, ti(]ii the hegiiicnt of lliu diu.mets;r udjucent to the chord. y Google Pkopo-^itiox XXVIII. Theorem 346. In a rigid inawjli-, the square of the hypol'-na^i equal to the sum of tlir K(;'(«ivx of the Imjs. Given the right trl ,IC thi' liypot^TiuPfl. Proof. Draw the line BF ± AC. Tlieo AC : AB^Ali : AF :.AC X AF= AI?. Arts Also AC: nC=nC : F<: .-. AC X FC=BC^. Adding equals, At' (AF-+ FC) ^Ti? -rluf . Or 'aJ^ = AJr + luf. 347. Cor. 1. 7n a right triang!'', the square of '■ilher leg U equal to the sqi^are of the hi/poti-nuae mliii<.s the aquari' of the other leg. 348. Cor. 2. In the square ADCf>. the diagonal di- vides the square into two right ti-iangles. Henee AC- = AB' + 'bC'=2 AJT. .-. AC^^ABV2, '>r~ = '~- Hence flte diagonal and the side of a square are incommtnsurahie. Ex. 1. If the legs of a Ei. 2. In tiie figure o t. A are 1 J- in. and a in., liiid the liypotenuaB. p. 204, show that ~AB- : BC^ = dF : FC. y Google _Vi;MEIiICAL fROlEiri'IKS -07 Proposition XXIX. Theorem 849. In any oblique triangle, tlte square of a side oppo- xUp an acute angle is equal to the sum of the sguares of the other two sides, diminished by twice the product of one of those sides hj the projection of the other side upon U. Given neiite ^ (7 in A ABC, and DC tlie iii-ojet the side BC on the side A C. To prove lB-=BC^+AXf — 2 AC X DC. Proof. If P fulls oil AC (Fi^. 1), AI)==AC—r)C. If Zt falls on ilCpi-odueed (Pig. 2), AD^DC—AC. In either case, Air^AC^ + IHf- — 2 ACX DC. ax. i. Adding Blf to each of these equals, I7r + £Z*'=l6'' + 7*t'' + !B7r— 24CX dc.k^.i. But, in thei-t. A ABD, Air + 1^)^ = ~AB" , An, -.m, and, in the rt. ABBCIKr+Bir^'BC^. (Wbyf) Substituting these values in the above equality. AB^ = mf + ACP — 2 AC X BC. ax, 8. 1, A line 10 i ind tlio projeetioi Q. E. K. Ex. line; i' [1. loii« ninkes on nnsle of 45" with a seeond ;i of tli6 first line on the senoiid Ex. 2. Find the 9a me, if the angle i» tiO°. tion ot 3, If the aide i 1 the base. jf an equilateral triivngle is a. find it'f projeo- Sx. 4. If, in I'ig. 1 , iX'=10, AC = 1':, ttud 6C = ^d% lliid.-!«. y Google PliOl^UWlTION XXX. 350. In any obtuse triangle, tJie square of the side oppo- site OH ohtuse angle is equal to the sum of the squares of ike other two sides, increased hy twice the product of one of the ' ' by the projection of the other side «j)oh that side. sides Given the obtuse lAGB iu the A ABC, and CJ) the projection of BG on AC produced. To prove AB^= Jo*+"EG^+ 2 AC X CD. Proof. AD=AC+ CD. As. 6 .-. A3'='Ad'+'CB^+2ACXC}). A^. i. Adding BI> to each of these equals, AI)~-\-inf=^^+'CJJr+'BD^+ 2 ACX CD. Ax. 2. But, ill the rt. A ABD, AI?+ lw^= AB'-. (Why f) And, ill the rt. A CBD, CI?+ 'bd'= 'BCT. (WLy •<) Substituting these values in the above equality, 'A}?'='A<f+'B<f-\-2 ACY. CD. Ax. 8. Q. Z. S. 361 . Cor. // the square on one side of a tnangle equals the sum of the squares ott the other two sides, the angle oppo- site the first side is a right angle; for it cannot be aeute (Art. 349), or obtuse (Art. 350). Et. 1. If, in the above ."igure, BC=10, AC = 2, and IB€A = 120'', find AB. El. 2. UAB = 2I}, ISC=U, iind AC = 12, find CD. y Google NTiMERIGAL I'EOPEItTlES 209 Proposition SXXI. Theorem 352. J/, in any triangle, a median be drawn to one side, I, The sum of the squares of the other two sides is egual to twice iJie square of half the given side, increased hj twice the square of the viedian upon that side ; and II. The difference of the squares of the other two sides is equal to twice the product of the given side by the projection of the median upon that side. Given the A ABC, AB>AO, AM tlie median upon BC, and MF the projection of AM on BC. To prove I. Air+ lc^=2 32?+ 2 'AM^. 11. Is — Ic^=2 BC X MF. Proof. In the A BMA and AMC, BM^SIG. AM^AM, &ndAB>AC. Hjp. .-. ZAMB k gresiter thun /-AMC. Art. 108. ,". ZAMB is obtuse {for His greater ilian half a straight Z), luohtme £S ABM, A^ = In acute A ACM, AG^.^MO^+AM^— 2 . Adding, Jb'+Ic'=2 BM''+ 2 AM'', {forMC=im). Subtracting, Tlf-^AX? = 2BCX MF. Ifor BM+MC=BC). y Google PLANE G eOJlETRY 353. Formula for median of a tiuangle in terms of its sides. Ill the Fig., p. 209, deiiothij; AB by c, AC by b, BG by a, and AM by )», by Art. 352, b'^ -\- 'j^ ^ 2 m^ -K?)% '■^(/y^ + . 354. If iu-o chords in i the segments uf one chord i inmts of the other chord. XXXII. TUEOEKM a circle intersect, the product of rqiial to the product of the seg- Given the O ADBC with Uie chords AB and CD inter- eectics at the point F. To prove AF X FB= OF X FJ). Proof. Dm y AD and GB. Then, in the & AFn ami CFB. lA- IC. (mk/i hcinfj jNeHSMra? hij i it:-/: IHI). Art. 2r,». Also II)=IB. (Wlyt) Hcuoe the & AFD and C FB are similar. (Why f) .-. AF: CF.FD: FB. {Why I) And AFXFB=CFX FD. (Wbjt) Q. £. S. 855. Cor. 1. If through a fixed point within a circle a chord be drawn, the product of the segments of the chord is constant, in whatever direction the chord be drawn. y Google PROPORTIONAL LINES 211 356. Def. Four directly proportional quantities are foar quantities in proportion iu such a way tliat both, the ante- cedents, bolong to, one figure and both the consequents to another figure. Four reciprocally proportional quantities are four quanti- ties in proportion in such a way that the means belong to one figure and the extremes to another figure. 357. Cob. 2. The, segments of two chords intersecting in a circle are reciprociUy proporlional (tiie segments be- ing considered us parts of the chords, not of the A). Proposition XXSIII. Theorem 358. If, front a given point, a secant and a tangent be drawn to a circle, the tangent is the mean proportional he- tween the whole secant and its external segment. Given AB, a tangent, and AC, a secant, to the cLrele BCF, and AF Vad external segment of the secant. To prove AG : AB^ AB -. AF. Proof. Draw the ehords BC and BF. Then, iu the A ABC and ABF, Isi ^ lA. Ident. IC= I ABF. Arts. 258, 264. .-. the ^ ABC and ABF are .'^iiiiihu-. (Why n .■. AC : AB = AB : AF. (Why !) yGoosle J-U HOOK III. PLANK GF.OMKTKY SbQ. Cor. 1. If, from a given point, a tangent and a secant he drawn to a circle, the product of the wJiole secant and its external segment is equal to the square of the tangent. 860. Cor. 2. If, from a gi fen point without a circle, a secant he drawn, the product of the secant mul its external segment is constant, in witatever direction the secant be drawn. Fop the product of each secant and its external seg- ment eqnnls the square of the tang'ent, which is constant. 361. Cor. 3. If two secants be drairii from an exter- nal point to a circle, the whole secants and their external segments are reciprocally proportional. pROPOPiTiox XXXIV. Theorem 362. The square of the bisector of an angle of a triangle is equal to the, product of the sides forming the angle, dimin- ished by the product of the segments of the third side formed h'j the bisector. Given the A ABC, CF {or t) the bisector of ZACB, aad m and n the segments of AB formed by CF. To prove t^ — ab — nin. Circumscribe a © about the A ABC, y Google KUMEKICAL PISOT'EETIEa Proof. Prbduee OF to meet the circumference at H",.aud di-aw the chord BH. Then, iu the A ACF mdi CHB, AA = III. (Why ?) And IACF= IHCn. (Why !) Hence the A ACF and CUB are similar. (Wby?) .-. 6: £ = ic + (:a. (Why ?) .-. ( {xArt)=al. (Why!) Or (a- + /2 = ai. Subtracting ix, f = (ih — tx. Ax. 3. But /;c = WH. Art. 3r)4. Substituting liin for te, 363. Formula for bisector of an angle of a triangle. m:n = h :«(Art.332).-.m + K:m = i + (( ; (.{Art. 309), or c ; m — o■■^- h -. h (Ax. 8^ he In like manner K^^ — —r- Substituting for i (I + b nhr^ _oh(f< +h + r){a +h- (^=--ab- Ei. 1. Oil the figure, p. 210, lot ^F=10, rR = 4, and FC^S. 'mil FI>. Ex. 2. On thfi flfc-urH, p. 211, let AC = IG ui.d .JC = ia, Find AF. Ex. 3. On the figuvo, p. 213, let 4C=1G, CB = 12, and 4C = 14. y Google Zli JiOOK in. PLANE GEOMETHY Pkopositiox XXXV. Tuboreji 364. Jn any triauyle, the product of any two sides is equal to the product of the diameter of the circumscribed cir- cle by the altiiitde upon the third side. Given the A ABC, ABCJ) a circumscribed O, BDthe diameter of this O, and Bi^'the altitude upon AC. To prove AR X BC=BD X BF. Proof. Draw the chord DC. Then, in the A ABF and DBG, I A = ID. CWhyS) IDCB is art. Z. (Whyf) .-. & ATJJi'and 7>/J6'are simihir. (Why!) .-. AB : BB^BF: BG. (Wl.yf) .-. A B X BC=BD X BF. (Why?) q. E. D. 365. Cor. The diameter of a circle circmnscribed about a triangle eqiiah the product of iivo sides of the triangle divided by the altitude upon the third side. Ex. In the above figure, if AB = 8, CC = U, and ilF = 4^, fled the rtidiua o£ tbe eireumBcribed (.'iicle. y Google construction problems 215 cohstroction problems Proposition XSXVI, Problem 6. To construct a fcurth proportional to three given Given the lines m, n, p. To construct a fourth proportional to «s, n, iuid p. Construction. Take auy two liuea, .4/* and AQ, making any convenient /.A. On J-Ptake J.B = »k, and BC^n. On AQ take AI)= p. Draw BB * Through the point C draw CB\\J)n, and meeting A^ at E. Art. 2T9. Then I>Ii is the fourth proportional required. Proof. An : BC=AI> : DR. Avt. :siT. Or m : n^p : DU. Ax, S. Proposition XXXVII. Problem 367. To consiriiet a third proportional to two given lines. Let the pupi! supply the construetion and pr'oof . [SUQ. Use the method of Art. :i6C, ninking p = n.] Ex. 1. Construct a fourth proportional to three lioea, j, 1 and I In. long. Ex, 2. CocBtnict a third proportional to two lines, 2 and 14 in. long. y Google BOOK m. PLANE GEOJIETliif Proposition SXXVIII. Problem To construct a mean proportional between iivo given Given the lines m and n. To construct a mean proportional between m anil n. Construction. On the line AP take AB= m, and BC= At B erect a ± to AG meeting the semi-eironinferenee at R. Art. 274. Then BE is the mean proportional reqnired. Proof. AB : BE=BB : BG, Art. 343. {Ihc J- to the diameter from atiij points in the eirciniferance of a circle is a mean proportioiial belweeii the segmenls of the diamvler). Substitnting for AB and BG their values »« and n, m : BB^BK -. n. Ax. 8. Q. E. F. Ex. 1. Construct the third propovtloaal to two I'mes, 1 and IJ in. long. Ex. 2. Construct a mean proportional between two lines, 1 and 2 in. long. Bi, 3. Taking any line as 1, oonstiniot j/S. y Google CONSTRUCTION fROELEMS 2l7 Peoposition XXXIX. Peorlem 369. To divide a givm straight line into parts propor- tional to a number of given liiies. "^' \ Given the straight lines AB, m, n aiiJ /!. To divide AB into parts proportional to »t, n and j). Construction. Draw the line AP, making any convenient angle with AB. On APmark off AC=»H, CD=n,mdI)F=p. Draw-BP. Throngh the points C and D draw lines || BF, and meet- ing AB at E and 8. Art. 279, Tliou AR, US and SB are the segments required. _ , AB ES SB ^"°^- AC^cB^W ^''■'''- {if two lines are cut by a wimbey ofparallets, the corresponding segments are iiroportional) . For A C, CD and BF substitute their equals »», n and p. „, AB B8 SB Then — = — = —-. ax. h, m 11 2) Q. E, If. 370. Def. a straight line divided in extreme and mean ratio is a straight line divided into two segments such that one of the segments is a moan proportional between the whole line and tiie other segment. Ea, Divide a given line lato parts prupc y Google 18 BOOK III. PLANE GEOMETRY Proposition XL. Problem 371. To divide a given straiglii line in extreme and, Q 'I Given the line AB. To divide AB in extreme and mean ratio. Construction. At one end of the given line, as B, con struct a X OB eqnul to \AB. Arta. 274, 275, From as a center and with OB as a radius, describe Sl circumference. Draw AO meeting the circumference at C, and produce it to meet the circumference again at F. On A B mark off AP eqnal to A€\ on BA produced take A<J = AF. Then AB is divided in extreme and mean ratio inter- nally at P, and externally at Q. Proof. 1. AF: AB=-AB : AC. ah. n5fi. .-. AF~An:AB=AB — AG:AG. (Win-?) But CF=20B = AB, and AG=AP. Hence AF~AB=AP, and AB — AC^PB. .: AP:AB = P£':AP,orAB:AP=AP: PB. As. 8, Art. 308, 2. AF: AB = AB : AC. (Why?) .-. AF+AB:AF=AB + AC:AB. (Why!) But AF+AB=QB. anH AB + AC^AF^QA. :. QB: QA = QA : AB. Ax. B. Q. £. F» y Google consteuction problems 219 Proposition XLI. Problem S72. Upon a given straight line, to construct a polygon similar to a given polygon, and similarly placed. Given the polygon ABODE and tlie line A'B'. To construct on A'B' a polygon similar to ABODE and similarly placud. Construction. In the given polygon, draw the diagonals AG and AD, dividing the polygon into triangles. At /i'^on the line A'/f, eonstruet ZA'B'O' equal to /.B; and at A' construct Zll'A'O" eqm\\ to ZBAO. Art. 2TS. Pi-oauce the lines B'C and A'C to raeiit at C. In like manner, on A'O' construct A A'C'D', equiangular with A AOD and similarly placed; au.l on A'D' construct the A A'D'E', equiangular with A ADE uud similarly placed. Then A'B'C'D'E' is the polygon required. Proof. The ^ A'B'C, A'C'D', etc., are similar to the & ABC, ACD, etc, respectively. Art. n24. Hence the polygons A'B'C l>' l-l' and ABCBE are similar. Art. :iJO, y Google PLANE GEOMETKV EXERCISES. CROUP SO SDllLAR TiilANGLES Let the pupil make a list of all the conditions that ? tivo triangles similar (see Arts. 323, 324, 325, etc). Ex. 1. Given AD ± BC, and BF 1. AC; prove & ADC and BFG similar. Ex. 2. In the samG figure, prove the and BFC similar. What other triangle figare i3 similar to A BFC f Ex. 3. Two isosceles triangles are si theii uqual Ex. 4. Two iBOBGeles triangles are Bimi equals a base angle of the other. Es. 5. Given arc -iC=aro J!C; prove A AFC and AFC similar. Ex. 6, Prove that the diagonals and bases of a trapezoid together form a pair of similar triangles. Ex. 7. AB is the diameter of a circle, BD is a tangent, and AD intersects the clreumfetenco angles ABE and ADB similar. Ex. 8. liC is a chord in a circle, AQ is the di.imeter perpendicular to BC and meeting it at 2f; AF is any chord intersecting BC in M. Pi'ove the A JJfif and APQ similar. Ex, 9. The triangle ABC is inscritied in a circle; the bisector of the angle A meets BC in 1) and the circumference in F. Prove the triangles BAV and AFC similar. Ex. 10. A pair of homologous medians divide two similar triangles into triangles which are similar each to each. Ex. 11. Two rectangles are similar if two adjacent sides of one are proportional to the homologous sides of the otlier. Ex. 12. Two circles intersect in the points A and B. AC and AD are each a tangent in one circle and a chord in the other. Prove the iiJ^C and JBi) similar. [8ufl. Prove IBAD= ^ACB, tti!.] y Google EXEBCISES. PROPORTIONAL LINES 221 373. Proof that lines are proportional. In order to prove that certain lines are proportional, or have propor- tional relations, it is usually best to show that the given lines are homologous sides of similar triangles. Sometimes, however, other methods of proof are used (as the theorems of Arts. 354 and 358) ; but these, if in- vestigated, are usually found to be the method of similar triangles in disguise. EXERCISES. CROUP 31 PROPORTIONAL LINES Ex. 1. Ontho iignpe of Ex, 1, p. 220, prove JI'X /JC=B-FX--ir', and TSC'X Or> = BOXFC. Ex. 2. On the figure of F.k. 5, p. 320, prote CP: CA=CA : CF. (Hence as P moves tho product of what two lines is constant i ) Ex. 3. The diagonals of a trapezoid divide each other into pro- portional eegments. Ex. 4. In the isosceles triangle ABC, AJi = AC,,oa tho side ^fl, the point P is taken so that PC equals the base. Prove AB X PB = m?. Ex. 5. In a triangle the median to the base bisects all lines parullel to the base and terminated by tlie sides. Ex. 6. If PQ is nnj- line throngli F. tl.e midpoint of the line AB, and AI' mid BQ are perpendicular to PQ, show thai the ratio PF: FQ Ex, 7. The triangle ^CC is inscribed in a circle. J^is the midpoint of the arc JC and Bi*" intersects the line 46' in ffi Prove AB : liO = AE: EG. [SUG. I'se Art. 332,] Ex. 8. If two circles intersect, the common ehoid, if produced, bisects the common tangent. [SUG. Use Art. 358.] Ex. 9. If two circles intersect, tangents drawn to the two circles from any point iu the common ishord produced ure equal. y Google 22'2 BOOK III. PLANE GEOMETEY Ex. 10. Givi'Ti AJ>. I'T and EG I ; prove PQ = UT. [SuG. Show that ^= If', by show- ing them equal to a uommon vatio.] Ex. II. Lilies aveiltawn from a point within the triangle, to the vertices of llis triangle AltC. From B' any point in OB, £'A' is drawn parallel to J)'J and meeting OA in A', aud B'C' is drawn parallel to BC and meeting OC in C Prove A'li' : AB=B'C' : BC, and the ti'iangles ABC and A'D'ff similar. Ex. 12. GiyenJ7(c;Dn£I7,and ^ ^ F any point in BC produced | prove AE' = IIQXI1P- [St'G. Compare the similar A ABR and BQD ^ also the similar & AMD and EBP.'] " " eXEFtCISES. GROUP 85 NUMF.RICAL PROPERTIES OP LINES Ex. 1. If AD is the altitude of the triangle ABC, Tl^—lc?^ luP'—m:"- . Ex. 2. If the liingoiials of a quadrilateral are perpendicular to each other, the sura of the squ.ares of one pair of opposite sides equals the sum of the squares of the other pair of aides. Ex. 3. The square of the altitvido of an equilateral triangle is three-fourths the square of one side. Ex. 4. If AB is the hj-potenusc of a right triangle, and the leg BG is hisected at K, ZZJ" - AM? = SOff-. Ex, 5, PlJ isaline parallel totbehypotenuse^J of arighttriangle ^JJC, and meeting JCiuPand^Cine, Prove Z^^ + 5P = Zb^+P^2. Ex. 6. Ill the right triangle AhC, BE and Cf bisect the legs AG naAdB in the points £ and F. Pvove iR^ + ici^=5B&. y Google PJ^- -PC^ EXERCISES. AUXILIARY LINES EXERCISES. CROUP SS ArXILIARY LIKES iiven ^Br/i ft veutHnglo; prove = 'PI'?-\-'PD-. Ex. 2. The eles divides tlie line of centers into segments which have the same ratio as the diameters ol the oircleE. [Suo. Draw radii to the points of coi Ex.3. n altitiidi ■IB and -■IC are the liigs of a Prove 2JCX-F'C = i(6'2. Ex. 4. Given the chords AB and CD perpen- dicular to each other and intersecting at O; prove (W^ + OB^ -r OC^ + 0O^= (diameter)^, [SuG. Draw tlie diameter BE and the chords AC, BD, DE. Prove AC=ED. etc] Ex, 5, ABC is an inacrihed i of which AB and AC are the legsi Ftov6A^=ADXAE. AD is a chord r 3elos triangle ABC, and D is n the t 6 produced, then CD^=CB- + ADX BI). Ex. 7. Two circles touch at the point T. I'TI" and QTO" are lit drawn meeting the circumferences in P, Q and P", (/ resiiective Prove the triangles PTQ and F'TQ' similar, [Sua. Draw the common tangent at T.] Ex. 8. Two circles touch at the point T, throngh 3" three lines i drawn meeting the circumferences in i', Q, E and P', (/, E', reapi tively. Prove the triangles PQR and PQ'Ji' similar. Ex. 9, If A is the midpoint of CD. an arc of a circle, and AP any chord intersecting the chord CV In Q, prove that APKAQ h y Google 2^i BOOK ill. PLANE GEOMET Ex. 10. In nn iiiseribed qnadriluterdl, the product of the dingonala ia equal to the sum o£ the products of the opposite sides. [8UG. Drfiw BFfO that ICBF= lABD and use similar trian{|;lts.] Ex. 11. The sum of the siiiian's of t]iy aides of any quadrilateral is equal to the sum of the squares of the diagonals, plus four times the square of the liue joining the loidpoiuta of the diagonals, EXERCISES. CROUP 34 INDIRECT DEJIOXSTRATIONS Ex. 1. If the sum of the squares on two sides of a triangle is greater than the square on the third side, the angle included by the two given sides is an aeute angle. Ex. 2. If D ia a point in the side AC of the triangle AJ!C. ajid AD : DC = AB : BC, then Dli biseeta angle ABC. Ex. 3. A given stmight line can be divided in a gtveu rritio at but one point. Ex. 4. If the sides of two triangles are par- allel, each to eauh, and a straight line be paasiid through each pair of homologous vertices, these lines, if produced, will meet Ex. 5. If each of three c: other two, the three common chords intersect in EXERCISES. CROUP 35 THEOREMS PROVED BY VARIOUS METHODS Ex. 1, In the figure on p, 204, show that ABy, BF=BCX AF. Ex. 2. In the same figure, if J'C=3vli'', sliowthatZS^ : fl(?- = ! : 3. Ex. 3. AB is the diameter of a circle and PB is a tangent. If AP meets the ciccumfeteneo in the point §, prove that APX. AQ = AB\ y Google MIfiCELI.ANEOTIS EXERCISES. THEOREMS Ex. 4i. If the line hisocting the parallel sides of a tra,pezoid b produced, it meets the legs produced in a common point. [SUG. See Art, 340.] Ex. B. In similar triangles, homologous uiediaua have the aam ratio as iiomologoua sides. Ex. 6. A diameter AB is produced to the point C : CI' is perper dicular to JC; PB produced meets the eiruumterence at Q. Prov the triangles A QB ai.d POP. similar. Ex. 7. If PA and PB are chords in a eirele, and CD is a line pai aliel to the tangent at J* and meeting PA and PJl at C and I), the triangles PAIS and PCD are similar. Ex. S. Given AB a diameter and AD and BO tangents, JCand DB intersecting ut any point F on the circumference; prove AB a mean propor- tional between the aides AD and BC. Ex. B. In any isosoelea triangle, the square of one of the legs equals the square on a line drawn from the vertex to any point of the base plus the product of the segments of the base. Ex. 10. A line drawn through the intersection of the diagonals of a trapezoid parallel to the bases and terminated by tho legs is bisected by the diagonals. [SuG. SeeEs. 10, p. 222.] Ex. 11, If a chord is bisected by another ehorl eath segmi the first chord is a mean propoitiouil letweei the begui'nts of th second chord. of Ex. 13. If two circle'* aie tangent exteinallj and a linn is di: through the point of lunta t terminated bv the cireumferoncea ohordu intercepted in the two circles are to ^ath other ai the rain Ex. 14. Thz-ee times the sum of the squares of the sides oE a Bugle equals (our times the sum of the squares of the medians. y Google 2'^(> BOOK ni. PrANK El. 15. Find the loeua of the jHklpoiuts of lines in a trinngle parallel to the biis" and ter- minated by the sides. Ex. 16. Given AB the iliam^ai-r. AP. PQIi, BR, tangents; pmve P^ X Qli a<:ousUnt(= ra- dius squared j, Ex. 17. is the center of h circle and A is any point within the circle ; OA is produced to J), so that OA'KOB equals the radius squared. If P is any point in the oivcumferonee, the angles 0J>^ and OBI* are equal, [SUG. Use Art. 327.] Ex. 18. Given AF=FI1, and ri{ \\ Jl! : prove BP : FP=HK : FK. [SuQ. EP : FP=UII: FH, utc] Ex. 19. If from any point P within the triangle AJIC the perpen- diculars /'§, ^.B, PTare drawn to the sides Jii,JC, BC, re^ p. actively, EXERCISES. QROUP 3a Ex. 1. Construct J- = — ; ulso .c= — . Ex, 2. Construct i- = \/ <i'— iF~, i. a., l/'lTT'f Th"-^. Ex. 3. Construct i = V''iai, i. B., "|/(3 u) /i. Ex. 4. Construct ■x = \/d' — t>r', i.e., 1^ «" - {l/fiTo'. Ex. B. Given a line denoted by 1, eoustiuct i/3; also iv/5. Ex. 6. Divide a line inta three parts proportional to 2, \, %, Ex. 7. Divide a line harmouieaily in the ratio 3 : 5, y Google KXEKOISES. PROBLEMS !2'i7 ti'iangie into segments propoi'tioual la Ex. 9, Divide aline into segmenta in the ratio 1 ; -j/S. Ek, 10. Given a point P in the aide JB o£ a triangle ARC, draw a line from P to AC produeeii so that the line drawu niiiy be bisected byBC. [SUQ. Suppose the required line, Pqp., drawn meeting liC in Q andJCinB. From P draw Pi H JC. Compare the APL^ and yjiC] Ex. II. Through a given point F in the are subtended by the chord AB draw a ohord which shall be bisected by AB. [Suo. Suppose the required chord drawn, viz., PQR. Jointlieeeoter O with P and y. What kind of an angle is OQP, etc,?] Ex. 12. Ill an obtuse triangle draw a line from the vertex of ths obtuse angle to the opposite Bide which shall be a mean pioporticmal between the segments of the opposite side. [Sua. Cireumscribe a eirole about the triangle and reduce the problem to the preceding Ex.] Ex. 13. Find a point F in the arc subtended by the chord AB such that chord PA : chord PB = 2 : 3. [SuG. Suppose the required construction made, and also the ehnrd .^B divided in the ratio 2 ;3 at the point Q. How do the angles Ji'g and QPB compared] Ex. 14. Given the perimeter, conatruet a triangle similar lo a given triangle. Ex. 15. Given tliH altitude of a triangle, construct a triangle Ex. 16. In a givun circle inscribe a triangle similar to a given Ex. 17. About a given circle circumscribe a triangle similar to a given triangle. Ex. J8. By drawing a line parallel to one of the sides of a given rectiiugle, dividt thu rcclanaio imu Iwo ainjilar rectangles. y Google 228 BOOK Hi. PLANE (lEOMETUY Ex. 19. InKoribfl a square in a given triangle. [Si'O. If ABC is the given tnaagle, sup- pose hOFE the required inaeribtd square, .ioin HE and produce it to meet AH \\ IIC. VroveJH=JK, etc.] 374. The metJiod of similars in solving geometrical problems is best shown by the aid of Ex. In the side JSC of a triangle JHt tlie perpendiculars from it to tho othBr sides shall be in the ratio 3 : 1. CoNsTRUCTiOH. At any point /' in J'' erect ft X PQ of any convenient length, lu a direction ± All draw Jfy= i Ql'. ioia.EP. V-coiaSiTaw tiT\\IlQ. Produce AT to D. Then D is the required point. Let the pupil supply tiio proof. EXERCISES. QROUP 3? PROBLEMS SOLVED BY METHOD OF SIJIILARS Ex. 1. In one side of a triangle find n point such tliat the perpen- diculars from it to the other two sides siial! be in the ratio «i : n. Ex. 2. Find a point the perpendiculars from which to tlio three sides of a given triangle shall be in a given ri [SUG. Use Ex. 1 twice.] Ex. 3. Construct a oircle which shall touch two givon lin thr. given point. [SUG. Let OA and OC bo the given lines and P the given point. Draw any OR toueliing the two liups {OB at r} and intersecting OPprodueed at X. Dtaw the chord A'F, etc.] Es. 4 Inscribe a. square in a given semi-circle. [Sua. Circumscribe a semi -circle about any given square, by taking the midpoint of the base of the square as a center, and the line from this midpoint to a uou-adjaeent vertex as a radius, etc] Ex. 5. Solve Ex. Ill, p 228, by the method of aimllara. y Google KXEKCISl'lS . PROBLEMS 229 375. Algebraic analysis of problems. The cmuUtions of a problem may ofitn be stated as an algebraic equation; by sohiiig the equation, the length of a desired line in terms of knotcH lines may then be obtained, and the problem solved by construeting the algebraic e^^ression thiis obinined. Ex. Find B point P in the line AB mah tLat 'Jf-= IBi-'. " Analysis amd Construction. Denotu AB by ; and PB\>ja — x. Tlien K' = 3{a — a;)'. Construct a v'3, whanee construct '— — ; lay ofE the line ob- tained, 88 AP, on AB; this gtv&s the point P of internal division. Similarly, the eonatriiction of - ^ — gives F' , the point of external 3ISES. CROUP 3S . SOI.VKD HY ALGI'IBBAIC ANALYSIS E. tbatJ/'-=2BP-. Ei. 1 . Find a point f in a given iii En. 2. Construat a riglit triangle, given one i a and the projection, 6, of the other k-g on i [SuG. Denote the projection of a on the hy| tennsebyai. Then o= = a: U' + 6), etc.] Ex. 3, Inscribe a square in a given aemicircle. Ex. 4p. From a given line out oft a part which shall he a mean proportional between the rumaiuder of the line and another given line. Ex. 5. Given AC and CB arcs of ilO°, and « a given line. Draw the ehoiii CQ intersecting AH in J'aothat7'y=(i. [Sto. .c=-j' = y'. <r.i: = (r-i'i,) (/--J), elc] / Ex. 6, Given the greater segment of a line di- y\C -i extreme and mean ratio, construct the line. y Google Zdii BOOK III. PLANE GEOMETTiY EXERCISES. CROUP 3Q PROBLEJIS SOLVED BY VARIOIIS .MF/I'HODS Ex. 1. Conatruet two lines, given tljeir Hum [a line AB) and their ratio {m ; «). Ex. 2. Construct two lines, given tiioiv diffovencK iind tlieir ratio. Ex. 8, Divide a trapeaoitt into two sitoilur tviifitKoids by drawing a liue parallel to the bases. [Suo. Coneeive the fisiire drawn, and compare the ratio of the bases in the two trapezoids formed.] Ex. 4. Construct a mean proportional between two given lines by Ex. 6. Prom a given point draw a seeant to a eircle so (hat the extprnal segment shall equal half the secant. [SuG. Draw a tangent to the O and use the algebraic method.] Ex. 7. From a given ©sternal point P, draw a secant meeting a circle in A and B bo that FA : AB= m : n. [SuG, Draw a tangent to the circle from the point P and denote its length by *. Denote PA by inx and ^S by lij-. Then in{m-\-ii)x' Ex. 8. Through a given point P draw a straight line so that the parts of it, incliKied between that point and perpendiculars dvawn to the line from two other given points, shall be in a given ratio. [SuG. Join the last two points, and divide tlie line between ttiem in the given ratio.] Ex. 9. Conatruet a straight line so that the perpendioulacB on it from three given points shall be in a given ratio. [Sue. Let F, Q, R, be the given points and m : n : p the given ratio. Divide FQ in the ratio m : n and Qlt in the ratio n : p, etc] Ex. 10. Upon a given line as hypotenuse construct a right tri- angle one leg of which shall be a mean proportional betwf - tlie other leg and the hypotenuse. y Google Book IV AREAS OF POLYGONS 376. A uait of surface is a square whose side is a unit of length, as a square inch, a square yard, or a square etutirneter. 377. TIip area of a surface is the number of units of surface which the given surface contains. It is important for tbe student to grasp firmly the fact that area menus not mere vague largeness of surface, but that it is a nainber. Beinf^ a number, it can be resolved into faKtors, it may be determined as a product of simpler numbers, and handled with ease and precision 378, Equivalent plane figures are plane figures having pqunl areas. Thus two triangles may have equal areas (he eqiiivaletit) and ret not be of the same shape, that is, not be equal (congruent). 379, Abbreviations. Instead of "area of a rectangle," for example, it Is often convenient to say simply "rectan- gle." So instead of "the number of linear units in the base," we use simply "the base." In like manner, for "product of the number of linear utiits in the base by the number of linear units in the altitude," a common ab- breviation is "product of the base by the altitude." (231) yGoosle -•J-i BOOK lY. PI,ANE GEOMETRY COMPARISON OF RECTANGLES rROPOSiTiON I. Theorem 380. If Itco rectimijles have the same altitude, llinj < to each other us their lases. Given the rectangles EFGJT and J7>CI>, having their altitiideri l-'Fand Ali equal. To prove EFGB. -. ABCJ)^ hW -. A 7). Case I. 'SVIien the hanes are (■ommensumNe. Proof. Take some eomraon measure nf fT/f and .1/), as AK, and let it be contained in EH n times and in Al) vi times. Hence EU -. AD = n im. (WLy?) Through the points of division of the hases of the two rectangles draw lines perpendicular to the bases. These lines will divide EG into h, and AC into m small rectangles, all equal. ^"- "'^' Hence EFGR : ABCD^m m. {Wh;?) .-. EFGS ■.ABCD = EU: AE. (Why!) yGoosle COMPARISON OF RECTANGLES 233 Case II. When ike bases are incommensurable. A D E Qil Proof. Divide the base AD into Jiny numher of ftiiinL parts, and apply one of these parts to EU. It will be contained in EH a certain nnmber of times ■with a remainder QU, less than the unit of measure. Draw gP ± EH, meeting FG at P. Then EQ and AI> are commensurable. Oonstr. . EFPQ _EQ ABCD A3 If now the unit of measure be indefluiteSy diminis the line QH, which is less than the unit of measure, be indefinitely diminished. .-. A-(>^£'/fu--alimit; EFPQ^^EFGHi>.i.i^\im\L An EFPQ T^_,^„,„^ ^ „„,.ifl>,v wlch ^^^^ . Case I. 381. Cor. // two rechinijUs have e(piul bases, iheij i to each oiher ai> their ullilutks. y Google 264: ROOK IV. PLANK OEO^tF,TKy PliOl'OSITION II. TilliUUKM 3S2. Tl)it ureas of any two reciani/hs are lo aii:h other as till- proOiirta of their bases iij their altitudes. Given the rectangles E and R', iiaving the bases b auiI 6', and the altitudes « and «', respectively, _ R h X a ^'^''"'' R=VX7^' Proof. Constrnet a rectangle, N, liaviug iUs ba^^e equal to that of R, and its altitude equal to thai, of B'. Also R'' Taking the produet of the corresponding members of the two equalities, Q. E. D. Ex. J. Fiud the ratio o &re 12 X 8 iu. to that of o: Ex. 2. How many brieks, eacL 8X5 avementeOX 9"-* of a rectangle whose di will it take to y Google AKEAK OF POLYGONS JoO AREAS OF POLYGONS Proposition III. Theorem 3S3. The area of a rectangle is equal to the product of Us base by its aUiluile, ,m Given the reotanglefi, with a base containing b, and an altitude containing h nnits of linear meaanre. To prove area of B=b X h. Proof. Let fTbe a square each side of whicli contains ( unit of linear measure. Then t-'^is the unit of surface. ArL. 'sm. 1X1 = bXh. — is the ar aof B, {h<, deniHllmi »/(i™n) amiof Jf = = bXk. (See Art. 1.) Ex. 1. Find, audS ft. wide. Ex, 2. The ar Fiud the biwe. Ase of this theorem, the probiem of liuding tbe reduced to tlie simpler problem of meaaurinj; ons of the reclaiigle and taking their product. i square feet, the X of a reeUnglB is i ingle 8 yds. long ,s altitude is 5 ft. y Google PROTOblTION IV. TliEOHE.M 385. The area of a parallelofjmm if ''<juid to thi; product of its base by ils aUitude. Given fcho OJ ABCD with the baso AT) (douote.! by h) and the altitude DF (denoted by It). To prove area of ABCB = h X h. Proof. From ^i draw AK || DF. and meeting CB pro- duced, at K. Tlien AK ± GK. Art. 133. .■. AZFD is a rectangle witli base band altitude /(.(Why!) In the rt. A AKB and 1)FG, AB^BG. AK=BF. :. A AKB^A t)FG. To each of these equals add the figure AliFD; Then rectangle A7i'i^O oro ABVD. But area of the rectangle AEFI)=b X h. .-. nrcaa ABCD= h x /*■ Q.E.D. 386. Cor. 1. Parallelograms which have equal bases and equal altitudes are eqmvalent. 387. Cob. 2. Parallelograms which. have equal bases are to each other as their altitttdes; Parallelograms ichich have equal altitudes are to each other as their bases. 388. Cor. 3. Any two parallelograms are to each other as the products of their bases a7id alliiadtin. (WLy!) (Why 7) (WhyT) Ax. 2. Art. 383. (Why ?) y Google AEEAS OF POLYGONS 3d7 Proposition V. Theorem 889. The area of a triangle is equal to one-half the product of its base by its altitude. Given the A ARC with tlio base AC {denoted by i), and the altitude Fli (aenoted by /(). To prove area of A ABC^i h X Ji. Proof. Draw BD \\ AC, and CD ]] AB, forming the /Z7 ABDC. Then BO is a diagonal of HJ ABJ)C. :. A ABC^hCJ ABDC. Art. ir.fi. But areaC7A£DC=6 X ft. (Why?) .-. area A ABC=i 6 X ft. Ax. 5. Q. £. D. 890. Cob. 1. Triangles which have equal bases and equal altitudes {or which have equal bases and their vertices in a line parallel to the base) are equivalent. 391. Cor. 2. Triangles n-htch have equal bases are to . fach other as their altitudes; Triangles which have equal altitudes are to each other as their bases. 392. Cor. 3. Any two triangles an- to each other as the prodticts of their bases and altitudes. Ex. 1. Find the area ot a parallelogram whosa liuaa \s 'J tt. 8 ir.. and whose altitude is 2 ft. 3 in. Ex. 2. Piud th« altitudu of a trian^'lB wIiu^b iiica is 180 sq. hi. ana whose base is 1 ft. 3 in. y Google 2u8 BOOK IV. WANE (iC(IMi:'r]:V PRoi'osiTioN VI. Theorem 393. If a, h, c denote the sides of a tnuiigle opposite the angles A, B, C, respectively, and s = i {a + h -{- c) , the area of the triangle = V^s (s — a) (s — b) (s — c). Given the A ABC -withthe sides opposite A A, B and C, denoted by a, b and c, respectively, i (a + b -\- c) denoted by s, iiiid A an acute angle. To prove area A AB(7 = V^s {s — .i) (s-h) (.--^T Pioof. Draw the altitude BB and denote BB by h. Then a-^b- + c- — 2h X AB. Art. 349. .-. 2&X AB^b' + c^~<i:'. Axs. 2, ....«.-±|^, But h'-<'' — A-D'-=(c + Al>) U — AD) Art. 3< =1"+^ — )\' — 2F^; •''■ . (i. + e + (0 (t + c — a) (»+!.-(!) (g — J + c) , a+ 6 — c = 2s — 2c, etc. Hyp,, Ass, 4, ; y Google AREAS OF POLYOONy '2?i^ 4 6- 4 6- 2\/,{s-a)U-b)U-c) B\it area A ABC^h b X h. Art- :m. •■. area AAiiC-v'i. (>,■ — «) (s — fi) {s~c). Ax, a. q. £. T>. Proposition VII. Theorem 394. The area of a frapczotd is ei^nal In tm uf Hn hasen muUiplied by its altitude. Given the trapezoid ABCB with the bases AT) and BO (denoted by& and b'), and the altitude FB (denoted by h). To prove area of AB<7D=i {h + h') X h. Proof. Draw the diagonal HI). Then area of A AIW= h b X it. (Why t) And area of A BVD^i V X h. (Wliy?) Adding, area of AHCl) = h (W -^ b) h. (Why?) Q. E. D. 395. Cor. The urea of a trapezoid equals the product of the ■iiD'd'dii of the trapezoid by the allitiide. For the njediaii of a trajjuKoid equals one-half the sum of the bases (Art. 179). y Google 240 ,ANE GKO^CETKY S96. ycHOLiUM. The area of a polygon of four or more sides can usually be found in one of several ways; as, by dividing the polygon into triangles and lahhtg the sum of the. aiv.as of tkf. triangles; or, by draiving the loiige^tl diagonal of the polygon and drawing perpendiculars to this diago- •ual from the vertices which it does not meet, and obtaining . the smn of tlie- areas of the triangles and frupfzoids thus formed. e 5, 6, and 7 in. are IS and 10 iii.. Ex. 4. If tte area of a trapazoii] is 135, and i 18, find its altitude. y Google COMPAKISON OP I'OLVGONS comparisom of polygons Proposition VIII. Theorem 897. If two trianffles have an angle of one equal to an angle of the other, their areas are to each other as the pro- ducts of the sides includhuj the equal angles. Given the To prove , ABO and AJ)^' having Zi AABC__ABXAC in eommon. A ABF ADXAF Proof. Draw the line DC. Then the A ABC and ADC may be regarded &s having their bases in the line AU, and as having the uoinmon vertex C. A ABC A P. ■-■AAIW^AJy ^''■'''- (Why t) l tlie corresponding members of these equali- A ABC ^ ABXAG A ABF ADX AF y Google IIOOK IV. PLANE GEOMETKY 1'EOPUSITION IX. TlIEOHEM 398. The arms of Iwu mnUar irkiitykn ar. as the squares of unij two hwiiuloijous shies. Given the similiii- & ABC and A'B'C with AB and A'B' homologous sides. A .4 lic _ ~n? To prove — -~", — r=zz:- ^ AA'iJ'C" A'B'- Proof. Draw the homologous altitudes CF and C'F' . A ABC AB XCF AB ^ CF Then A A'B'O A'B'XC'F A'B' C'F' ' " arc to each other as the pymhicts of Ihfir hmes and aliitutlei)). CF A B Substituting TTTT/ foi' its equal 7^77^' A ABC ^ ^^ y^ --^J^ A A'B'V A'B' A'B' AB- ~VB'- Ex. ]. If a pair ot lioroologous siJes of two similar triangles 4 ft. aud 5 ft., find the ratio of the areas of the triangles, Ex, 2. Prove Prop. IS Ijj use of Prop. YILI. y Google compjIeison of polygons 2i6 Proposition X. Theohem 399. The areas of tivo similar polygons are to each other s the squares of any two homologous sides. Given tiie similar polygons ABCDE and A'B'C'D'E', ■with their areas denoted by S and S', respectively, and with AB and A'B' any pair of homologons sides. To prove S r S'=Xb^ : A'B''. Proof. Draw the diagonals AG, AD and A'C, A'B' from the homologons vertices A and A'. These diagonals will divide the polygons into similar & . Art. 32!l. A ABC AB^ . ^ Art. 308. _^A'B'G' A'B'"' " A^'B't" V7^^7 AA'Ci)' Vl^V AA'B'E' (Why?) A ABC ^ A AGB A ABB _ ■'■ A A'B'V £\A'GB' AA'B'B' (^'^'y'^ A ABC + A AC I) + A ABB __ A AUG _ ^^^ .^^„ ' A A'B'C'+ A A'G'B'-i- A A'B'E' A A'B'G' . '"^ ^ A ABC ^^ ,, •• «' A A'B'C" . ^ = J^. ^, 1 ^' -^I'-t;'' o. E. D. Ei. If a pair of honiologouH sides o! two site "at! 2 ft. , Cud the ratio of the areae o£ the polygoi y Google I'LAXE C.EOMKTKV Peoposition XI. Theorem 400. Til a right triangle, ilie square on the htjpotenu! is eqtdvaknl to tht sum of the squares on the two legn. Given AT> the square on AC the hypotenuse of the rt. A ABC, and BF and BK the squares on the leys Ba ;md BC respectively. To prove ADo BF + BK. Proof. Thi'oiigh B draw BL \\ AB, and meeting KB in B. Draw BE and FC. Then i ABC and ABG are rt. A . :. GBCls a straight line. Tn the A BAE and FAC, AB=^AF, i Also But (for AL hai Also ZBAB= IF AC, (for cach^ZBAC + arl. . :. A BAE=/\FAC. d AB=Aa. I. Why ?) In like Adding, square BF^2 A FAC. rectangle Ai— square BF. rectangle Ci~sqaare BK. AL + CL, or AB^BF+BK. As. : J A BAE). (Why ?) 401. Cor. 2'ke square on either leg of a right triangle is eqiiivalent to the nquare on the hypotenuse ditninished by the square on the other leg. y Google CONSTEUCTION I'iiOBLEMB COKSTKUCTIOK PROBLEMS Proposition XII. Problem 402. To construe/, a square equivalent to the sum of iwo given squnrex. c A Ir Given two squares S and S'. To construct a square oquiviilciit to S 4- •'^'- Construction. Constrnct a right angle BAC. ah. 574. On one side of this angle -take AB equal to a side of S, and on the other side take AG equal to a side of S'. Draw BC. On a line equal to BC construct the square B. Then B is the square required. Proof. B=BC^'^'Jb'^ + A(f. Arf.4r)0. .-. B=^S+ iS" Ax. 8. Q. E. F. 403. Cor. To construct a sqtiare e<inivule}i.l to the sum of three or more ghm squares. At C in the above figure erect a liue GJ) X BC (Art. 274), and equal to a side of the thii-d given square. Draw DB. DB will be a side of a square equivalent to the sum of three given squai'cs, etc. Ez- 1. CoBstruot a square equivalent to ths s ■whose sides are i in. and 1 in., respectively. Ex. 2. By use of Art. 40J, taking a given Um v'S; also v/'l. For example, eonstraet 1 o( t ) squares y Google ^'■iO liOOK lY. PLANE GKOMETET PiiOmsinoN Xin. 3'roblkm 404. Tomm'InicI a ^qiuirc equivalent (o the difference of two yiven aquare^i. B B Given the squares W aiul S'. To construct n sqnarij equivi and S'. Construction. Coustn hnit t.o iln; difTnrcncR of ; ■t'Ait&ngh BAR'. Art. 274, Ou one siilo J B take AB equal to a side of the smaller given square 8'. From B as a center with a radius BC, etjnal to a side of the larger square, describe an are intersecting AK in G. On a line eqnal to AC eoustruct the square R . Then B is the square required. Proof. B = AC-^BC- — AB', Art, 401. {the nqiiare on either leg of a right triangle is equitiateni to the square on the hypotenuse diminished by the square on the other leg). Ex. Construct fl square e whose sidtB are 1 in. and i ii Q. E. F, ilitTereuce of two aquarea y Google CONSTRUCTION PROBLEMS ^H Proposition XIV. Problem 405. To cimstriid a square eqxdmhnt to a given paml- lelogram. i) k -p Given tlie CJ AliCT) with base S and altitude h. To construct a sciuave equivalent to ABCD. CoEstructioa, On the lino EO take EF (sqnal to h and FG equal to 6. On EG iis a diameter construct a semicircle. Art. 27a, Post. 3. At F erect a X meeting the semieircumference at K. Art. 274. On a line equal to FK construct the square S. Then S is the required square. Proof. S^KF'. But KF''^bXh, Art. S43. (n ± from a«ij poiii! hi a cinyiimfcrnnw lo a i^iaJiii^lci- is n mean pyopor- tioiial iftwecn lite sajotenls of the diameter). But area CJ ABGD^b X h. (Why t) .-. 6'oarea £Z7 ABGD. Q. E. F. 406. Cor. To construct a square equivalent to a giien triangle, eonstriiet the mean proportional between the base , and half the altitude of the triangle and construct a square oa ihis mean proportional. y Google i8 BOOK IV. PLANE GEOMETRY Peoposition XV. Problem 407. To construct a triangle cqmvaleid to a give Given the polygon ABCBE. To construct a triangle equivalent to ABCDE. Construction. Let --i, B, (7 be iiny three consecutive ver- tices in the given polygon. Draw the diagonal AC. Draw BF \\ AC (Art. 279), and meeting AE produced at#. Draw FC. In the polygon FGDE take the three consecutive ver- tiees C, B, E, and draw the diagonal CE. Draw BG || CE, and meeting AE produced at G, Draw CG. Then A FCG is the triangle required. Proof. A ABC->AAFG, Art. wo. (having the same base AC, and their veriiees in a live BF |] the base) . Also A AGE = A ACE. Went. And A FGB = A ECG, Art. 390. [hamng the same base CEand their vertices in line DG [j >Mse). Adding, A ABC + A AGE + A ECB o A AFC + A.1CB+ A ECO. As- 2. Or polygon ABCDE<^ A FCG. »}. ffi. F. 408. Cor. To construct a square equivnUnt to a gi^en polygon, use Arts. 407 and 406. Ei. Construct a triangle equivalent to a given hesagon. y Google CONSTRUCTIOK PROBLEMS M\) Proposition XVI. Pkoblem 409. To construct a rectangle equivalent to a given square, and having the sum of its iase and altitude equal to a given line. r D. Given the sf|uiire S and the line ,4 B. To construct a rectangle equivalent to S, nnd having the sum of its base and altitude equal to AB. Construction, On AB as a diameter describe the semi- circumference ADB. Art, 275, I'OBt. 3. At the point A erect a X , AG, equal to a side of S. Art. 274. Through C draw a line || AB (Art. 279), and meeting the cLryumference at I). DrawDE J. AB, Art, 273. Construct the rectangle R with a base equal to EB and an altitude equal to AE. Then B is the rectangle required. Proof. DB- = AJ'JXEB. (Why f) But DE=CA. (Wby ?) .: CA^=^AEXEB. (Why?) Or S=>B. 410. Cor. The above problem is eqiiivaieiit to the problem: Given the sum and product of iivo lines, to con- struct the lines. y Google 50 BOOK IV. I'LANE GIilOMETKY Proposition XVII. PiioJiLEJi 411. To eon&iruct a rectangle equivalent to a given square, 'tid having ike difference of Us base and altitude equal to a liceu line. [ZZZl Given the square S and the line Ali. To construct a difference of its 1: ictangle equivalent to .S', and ie and altitude equ<il to AB. Construction. On AB as a diameter describe the c fereuce ABBF. Art. 275, Post. 3. At A erect the X AC equal to a side of S. Art. 274. Draw C'F through the center 0, and meeting the circum- ference at the points D and F. Construct a rectangle E with base equal to CF and alti- tude equal to CD. Tht:ii li is the rectangle required. Proof. CF: CA = CA : CB, Art. -isa. .: CA'=CFX CIK (Why !) :. S^B. (Why ?) Also the difference of the base and altitude of B= CF—CB = DF=AB. Q. E. F. . 412. Cor. The above problem is equivalent to tbo problem: Given ike difference and the product of two lines. to constrtKt the Ums, y Google CONSTRUCTION PROBLEMS ^Ol Proposition XVIII. Problem 413. To construct a polygon similar to two given simi- lar polygons, and equivalent to their sum. B' K"""L Given the similar polygons P and P'. To construct a polygon similar to P and P', and equiva- lent to their sum. Construction. Take any two homologous sides, AB and A!B', of P and P'. Draw J/A'X KL (Art. 274). making l/i:=J.B, and ^i = A'B'. Draw ML. On A"B", equal to ML, as a side homologous to AB construct the polygon P" similar to P. An, 3T2. Then P" is the polygon required. Proof. I" A«B«'' "'■" P" ^"B"'' Art. 399. (a. „r,a, o/e '1-rt similar polygons ore k> mch other a. their homologous siiles). , P+F' lYf + A^'^ ? Ihe *9"a'"i^» 0/ Atldiii?. As. a. ^P" ^ .1"P"^ But jU"2i:^"+ A'i^ = i¥//, Art, 346. Or AB--\- A'B'- 'a"B"^_i As. 8. As. 8. y Google BOOK IV. I'LASE GEOMETKY Proposition XIX. Proulem 414. To construct a square which shall have a given ratio (o a given square. B A'-^:[yW"";c Given the square S and the lines m a.nd n. To construct a square which shall be to S In tlie ratio Coastruction. Take AB equal to a side of S and draw AF, making a eonveuieut angle with AB. On AF take AD equal to m, and J>F equal to w. Draw DB. Draw F€ \\ DB, meeting AB produced in C. Art. 270. On AC, as a diameter, construct a semicircumference AKC. Art, 27-1, Post. 3. At B erect si L BK meeting the semieireumference at Tf. Art. 274. Construct a square B' having a side equal to BK, or x. Then &' is the square required. Proof. ^^ = a X 6. (Why?) Also a : h = m -. n. (Why?) □„„„„ «_«!_«!_"_'». ,,. . . S' ab b yGoosle OOKSTRUOTION PRORLEMa 2:)d Proposition XX . Pboelem 415. To construct a polygon similar to a given polyg&ii, ind having a given ratio to it. Given the polygon 7" and the lines m and n. To construct a polygon /-" which shuU be similar to P, and be to P in the ratio n : m. Construction. Construct a square which shu.!! be to clie square on AB as n : ni. Art. m. Let A'B' be a side of this square. Upon A'B' as a side homologous to AB construct a polygon P' similar to P. Art. 37'J. Then 1" is the polygon required. p_ AK I' 375 AJP -". JTb'- n y Google PLANK GEOMETIIY Pkoposition XXI. Problem 416. To construct a polygon similar to one given polygon ^ and iquiviiUnt to another given polygon. O^ Givea the polygons P and Q. To construct a polygon similar to P, and equivalent to Q. Construction. Conatruet a square equivalent to P, and let in tie one of its sides. Act. 408. Construct a square equivalent to Q, and let n be one of its sides. Art. 4oa Construct >1'/J', the fourth proportional to ni, n, and AB Art. 3GG On vl'-B', as a side homologous to ^l/>', euiistruet a poly gon P" similar to P. Art. a72, Then P is the polygon required. ^ , P m^ Tli^ P Proof. _= =.^^=-_. Constr , Arte. 314,399. P P As. 1. .-. F^Q. Art. 305. Q. I. F. y Google EXERCISES . THEOREMS 2 JO EXERCISES. CROUP 4«> THEOEF.>[S CONCERNING AREAS Ei. 1. The dia{;oiial3 of a parallelogram divide the parallelogram' into four equivalent tnangles, Ex. 2. Any straight line drawn through the point o£ intersection of the diagonals of u parallelogram divides the parallelogram into two equivalent parts. Ex, 3. If, in the triangle ABC, D aud F are the midpoints of the sides AJi and AC, respeetively, the area of ADF equals one-fourth the area of ABC. [SuQ, Use Art. 397,] Ex. 4. If the midpoints of two adja.'ent aides of a pruallologram be joined, the area of the triangle so fonned equals oiii^-eighth the area of the parallelogram. Ex. 5. If, in the triangle ABC, I) and /■■ are the midpoints of the sides AB and AC, respectively, the triangles ADC and AFB are equi- valent, Ex. 6. In a right triangle show, by obtaining expressions for the area of the figure, that the product of the legs equals the product of the hypotenuse by the altitude upon the hypotenuse. Ex. 7. If two triangles are equivalent, and the altitude of one is three times the altitude of the other, find the ratio of their bases. Ex 8. If two isosceles triangles have their legs equal, and half of the base of one equivalent to the altitude of the other, the tri- angles are equivalent. Ex. 9 If two triangles have an angle of one the supplement of an angle of the other, their areas are to each other as the products of the sides ineludinj; these angles. (I b Ex. 10. Prove geomiitrieally that (ii+ii) =" •!■'■ [ ~~1~ +2(ib. Ex, 11. Similarly prove («—i)'=a' + i''~2u6. Ex 12. Similarly prove (o+f)}(n—;>)=(i=—i/'. Ex. 13. The line joining the midpoints of the parallel aid^s of u trapezoid divides the tvupezoid into two i^iiuivitliini piirt^.. y Google ii56 "BOOK IV. TLANr; GKOMETUV Ex. 14. The linea joiuing tlio miiipolTit of otio diagonal ot i quadrilateral to the verticea not joined by the diagonal divide the quadrila,teral into B T O two equivalent parts. / / Ex. IB. Given QR and TS passins through P, any point on the diagonal AC Df a CJ , QR\\ AD, and TS ]] All, prove QBTP<:^PRDS. Ex. 16. Given OH^OD; proveAABC^ AADC. I-et the pupif also state this as a USE Of ACXLLIARY LINES Ex, 1. Given AJICI) a ZZ7 and P any point inside ABDC; prove APJD + APBG = £:.PAB+APCD. Ex, 2, The area of a triangle is equal to one halt the product o£ its perimeter by the radius of the inscribed Ex. 3 If the estremitiea of one leg of a trapenoid bo joined to the midpoint of the other leg, the middle one of the three triangles thus formed is equivalent to half the trapezoid. Ex. 4. Tho area«f a trapezoid is equal to the product of one leg by thn pi'rpeudicular on that leg from the midpoint of the other leg. Ex 5. If the midpoints of the aides of a quadri- lateral be joined in order, the parailelogiam thus formed is equivalent to one-half the quadrilatetal Ex. 6. A quadrilateral is equivalent to i triangle two ot whose aides are the diagonals of the q u id i: lateral, the angle included by these sides being equal to one of the angles formed by the intersection of the diagonals. y Google EXERCISES. THEOREMS J,i>i EXERCISES. CROUP 4S THEOEEMS PROVED BY VARIOUS METHODS Ex. 1. If through the midpoint of one leg of a trapeioid a line be drawn parallel to tde other leg to meet one base and the other base produced, the parallelogram so formed is equivalent to the trapezoid, Ex. 2, If the midpoints of two sides of a triangle be joined to any point in the base, tbo quadrilateral so formed is eqaivalent to halt the triangle. Ex. 3. If P is any point on AC the diagonal of a paralleloeram ABCD, the triangles AFB and APD are equivalent. Ex. 4. If the siili) of an equilateral triangle be denoted by a, tha area of the triangle equals -~r~- Ei. 5. Find the ratio of the areas of two equilateral triangles, i£ the altitude of one equals the side of the other. Ex. 6. If perpendiculars be drawn from any point within an equilateral triangle to the three sides, their sum is equal to the alti- tude of the triangle. Ex. 7. If -E is the intersection of the diagonals AC and SD of a quadrilateral, and the triangle ADE is equivalent to the A £EC, then the liues A IS and CD are parallel. Ex, 8. If, in the quadrilateral AliCD, the triangles AUC and AIiC are equivalent, the diagonal AC bisects the diagonal BD. Ex. 9. If two triangles have two sides of one equal to two sides of the other, and the included angles supplementary, the triangles are e(iuivalent. Ex. 10. II, in the parallelogram -4lJ?CD, F is the midpoint of the side BC, and AF intersects BD in K, the triangle liEF=-f^ the parallelogram ABCb. Ex. 11. Given PQ [[ AC, and PR \\ AB; ^ prove AQAB a mean proportional between AHQP and AFRC. Ex. 12. P Ik any point in the side llC of liie parallelogram A/!(-lr and HP pro- duced meets .1 /( produced iu <l Show that the triangles BPA and CPQ are equivalent. y Google 258 BOOK IV. PLANE GKOMRTIiY EXERCISES. GROUP 43 PROliLEMS IN CONSTRUCTING AREAS Ex. 1. Construct ii square having twii;*; Ihi; nrea i)£ n given square. Ex. 2. Construct a square hiiviug three times tho area of a given, square. Ex. 3. Construct a square oquivaient to tbe sum of three pivett squares. Ex. 4. Transform a given triangle info an equivalent isosceles triangle having the same base. Ex. 5. Transform a given triangle into an equivalent triangle having the same base, but having a given angle atjjaeeut to the base. Ex. 6, Transform a triangle into an equivalent triangle with the same base, but naving another given side. Ex. 7. Transform a parallelogram into an equivalent parallelo- gram having the same base, but ooutaining a given angle. Ex. S, Construct a triangle similar to a given triangle ami con- taining twice the area. To construct a similar triangle containing five times the area: how is the construction changed ! Ex 9. Bisect a given triangle by a line parallel to the b^se. Ex. 10. Constmet a polygon similar to tivo given i^imilar pohgons, and equivalent to their difierence. Ex.11. Draw a line parallel to one side of a given rectangle, and cutting off five-sevenths of the area. Ex.12. Bisect a parallelogram by a lino perpendicular to the base. Ex. 13. Through any given point c km , line U„Mme , t-iyea parallelogram. Ex. a. Construct a triangle equivalent to a given triangle, ABC, having a given base AD, but the ZBAG adjacent to the base un- changed. [SuQ, Draw CE \\ DU, etc.] y Google EXEHCISES. PEOBLEMS 259 B:i. 15. Transtorm a parftllelogram into an equivalent parallelo- giaxa haviagagivenbaee, but the angle adjacent to the base unchanged. Ex. 16. Tranaforra a given triangle into an equivalent riglit triangle having a given leg. [SUO, Uhb Ex. 14, then Ex. 5,] Ex, 17. Transform a given triangle into an equivalent right tri- angle having a given hypotenuse. [SuG. Piod the altitude upon tlie hypotenuse of the new triangle by fijsding the fourth propovtionnl to what three liiu's ? ] Ex. 18. Transform are-entrant pentagon into ; equivalent triangle. Ex. 19. TranBform a given tnaugle ioto equivalent equilateral triangle. [Sue. See Art. 416.] Ex. 20. Biseat a triangle by a line perpendicular to one of its aides. [Sue. See Art. 416,] Ex. 21. Construct a square equivalfiit to two-thivila of a given square . EXERCISES. CROUP 44 PROBLEMS SOLVED BY AEUEBHAIC ANALYSIS Ex. ]. Transform a given rectangle into an equivalent rectangle with a given base. Ex. 2. Transform a given square i; given leg. Ex. 3. TrauBform a given triangle right triangle. ato a right triangle having a into an equivalent iaosneiea Ex. i. Draw a line cutting off from triangle equivalent to one-half the give [SuG. Use Art. 3SIT.] , a given triangle an isosceles a triangle. Ex. 5. Through a given point in draw a line biaecting the area of the tri one side of a given tz'iangle angle. Ex. 6. Transform a giveu square into a rectangle which shall have three timeu th« jietimetec oi the given square. y Google Zbi) BOOK IV. PLANE GF.OMETHY EXERCISES. CROUP 4S PR0RLE3IS S0LVI3D BY VARIOrS >[ETHODS Ex, 1, Bisect a given, paralle log vara by a line parallel to the base. El. 2. Tmnsforni a pavallBlogr.am into an equivalent parallelo- gram having the aame base and a given side adjacent to the base. Ex. 3. Construct a square which shall aontain four-seventhg of the area of a given square. Ex.4. In two different ways i constr uct a square t .aving three times the areaof aglvensfjuare. Ex.6. Trisect a given triangl e by line IS piiraliel to thi i base. Ex. 6. Find a point ivithin a triangle such that line; i drawn from it to the \ 'erticestriseet the area. Ex. 7. Find a point within a triangle such that iinci •. drawn from it to tha three vertices divide the area i iito parts whic ■h shall have the ratio 1 2:3:4. [Sue. II one of the small A eontains i tlie area o£ . original A, a line thcoi igh its vertex cuts off s the altitude, etc.] Ex. 8. Divide a triangle ini to tbree equivalent parts by lines Ex. 9. Divide a triangle into three equivalent parts by lines drawn through a given point J' in one o£ the sides. [Suo. Use Art. 307.] Ex. 10. Divide a giveu qiiadrilatei-al into three equivalent parts by lines drawn through a given vertes. Ex. 11. Through a given point in tlie G base of atrapezoid draw a linebiseeting the ,' ''., area of the trapezoid. / \ Ex. 12. Bisect the area of a trapezoid y ' \ p by a line drawn parallel to the bases, p[ \ p [SUG. Construct A GKF similar to A J \ p ABG aud equivalent to i num el wliat two y Google Book V REGULAE POLYGONS. MEASUREMENT OF THE CIRCLE 417, Def, a regular polygon is a polygon tliat U both equihiteral aod equiangular. Proposition 1 . Theoreie 418. An equilateral poiyijnii (hat in in'^i-rihed in o circle is ulso eqaiaiigtilar and regular. Given ABC . . .K an inserilied polj^goii, with its sides at;, nc, CD, etc., etiiia!. To prove the polygon ABC...K equiangular and rogular. Proof. Arc A7.' = ai-c mj^avc- Cl>, ptc. Art. 21M. .-. are ABG^ayq. BCI}=arc CJ>E, etc. Ax. 2. .-. ^ ABC^IBCI)^ ^CDE, etc.. Art. 2C0. (all 1 inscribed in the same segment^ or in equal segments, are equal). .: tlie polygon ABC , . . Jt is equiangiiki'. /. the polygon ABC . . . KU regular, Art, 417. 0, z. B. {■JCI) yGoosle 262 KOOK V. PLANE GEOMF.TKY 419. Cor. 1. // the arcs suhtended by the sides of a regular insrrih''(l polygon he bisected, and each point of bisection lie joined to the nearest vertices of the polygon, a regular inscribed polygon of double the number of sides is formed. 420. Cob. 3. The perimeter of an inscribed polygon is less than the perimeter of an inscribed polygon of double ike number of sides. Proposition II. Tiif-orem 421. A circle may be circumscribed about, and a circle may be inscribed in, any regular polygon. Given the regular polygon ABCIIE. To prove that a O may be circi-imBeribeil about, or in- scribed in, ABCBE. Proof. I. Throueh A, B aad C (Fig. 1) , any three suc- cessive vertices of the polygon ABCDE, pass a circum- ference. Art. 235. Let be the center of this eircumferenee. Draw the radii OA, OB, 00. Also draw the line OD. Then, in A OBG, OB=OG. (Whj-r) :. lOBC^lOCB. (Whjf). yGoosle KEGULAE POLYGONS 263 But Z ABG = Z BCD, Art. 417. (being A of a regular j)Olii3on). Subtracting, Z OBA = Z 00i>. (Why?) Hence, iu the A OAli and OCD, OB=OG. (Whyt) ^B-Oi). (Why?) Z OBA = Z OC'D, (JH;;! jtniuecf). .-. A ABO^AOVB. (Why!) .-. 01>=0A. (Why?) Hence the cireiirafereuce whieli passes through the vertices A, B and C, will iilso pass throuirh the viirtes I). In like manner, it may bo proved that this circumference wdl pass tbrongh the vertex E. Hence a circle described with as a center, and OA as a radius, will be circumscribed about the given polygon. H. The sides of the polygon ABODE (Fig. 2) are equal chords iu the circle 0. Hence they are equidistant from the center. Art. 22(i. .■. a circle described with as a center, and the distance from to one of the aides of the polygon as a radius, will he inscribed in the given polygon. q. E. B. 422. Def. The center of a regular polygon is the com- mon center of the inscribed and circumscribed circles, as the point in the above figure, 423. Def, The radius of a regular polygon is the radius of the circumscribed circle, aa OA in the above figure. 424. T>FF. The apothem of a regular polygon is the radius uf tliL- inscribed circle. y Google 2fJ4 BOOK V. PLANE GEOMETRY 425. Def. The angle at the center of a regular polygon is the angle between two radii drawn to tlie extremities of any side, as the augle AOB. 426. COE. The angle id the. center of a re</Hlar polygon is equal to four right angles divided l»j the ntiinher of sides. Hence, if n denote the number of sides iu the polygon, the angle at the center of a regular jiolygon eijuulx —~ — ; also Ike angle between un apotkem and the nearest radius, in a regular polyjoit of ii sidvs, eqiiaU PHO^osITIO^f III. Theorem 427. If the circumference of a circle be dirided info any number of equal ares, I. The chords of these arcs form a regular inscribed polygon; II. Tangents to the circrimference at the points of di>:ision form a regtdar circumscribed polygon . e^:^c Given the circumference ABC, divided into the equal arcs AB, BC, GJ), etc., the chorda AB, BC. etc., and PQ, QB, etc., lines tangent to the circle at B, (J, etc. To prove ABODE a regular insei-ibcd polygon, and PQRST a regular circumscribed polygon. y Google eegUlak polygons Libu Proof. I. The chords AB, BC, CD, etc., are equal. (Why r) .■- po'ygoii ABODE is equilateral and regular. Art. 418. II. Ill the A APH, BQG, CRD, etc., AB=liC=Cn, etc. (-Why!) Also ZP.4B- ZPfiA = IQIiC^ IQGB= ^ BCD, etc. Art. 264. {«id< la:. .'. A AZ-'7i, /iX'<-''. Clilt, lite,., areeqiiiil, isosceles triangles. (Why?) .-. ^1'-^ IQ ^ Z7i*, etc. (Why?) Afld AP^PB^BQ=QC, etc. (WhyT) .-. PQ=QB = nS, etc. Ax. 4. .-. PQBST is a regular i)Cilygoa. Art. 417. Q. E. ». 428. Cor. 1. If the arcs AB, BC, CB, etc., be bisected, and (1 tangent fie draimi at each point of bisection , a circum- scribed regular polygon of double the number of sides of PQI18T will be formed. 429. Cor. 2. The perimeter of a circumscribed regvlar polygon is greater than that of a circumscribed regular poly- gon of double the number of sides. Ex, 1. Fiud the number o!^ <Jep:vees lu the central angle of a legalar poiitagon. Of a regular hexagou. Of a square. Ex. 2. WUat is the sliovt name for an iuscribeii equilateral quad- Tilu^al ; y Google 26G BOOK V. PLAVE GEOMETRY Proposition IV. Theorem 430. Tangents to a circle <at the midpabits of the arc? subtended hij the sides of a regular inscrihed pobj'jon form a regular circumscribed polygon whose sides U7'€ purulUl to the corresponding sides of the inscribed polygon. Given the regular polygon ABODE iiiscribed in tlie O ACD; P, Q, B, etc., tiie midpoints o£ tiie arcs AB, BO, CJ), etc.; and A'B', B'G', CD', etc., tangents to the circle at P, g, E, etc. To prove A'B'C'D'E' a regular polygon with its sides || corresponding sides of the polygon ABCDB. Prool The ares AB, BC, CD, etc., are equal. Ait. 218, .-. the arcs AF, FB, BQ, QG, etc., are equal. Ax. 5. .■, the arcs I'Q, Qli, BS, etc., are equal. Ax. 4. .■. A'B'C'D'E' is a regular polygon, Art. 427. {if Ike cii-ciimfereiiee of a O ho diiikkd, clc.). Side AB A OP. Art. us. A'B' ± X>P. (Whyt) .-. AB II A'B'. (Why?) In like manner, each pair of homologous sides in the two polvgons is parallel. Q. E. D. 431. Cob. Homologous radii of an inscribed and a eir- eumscribed regular polygon, wlhse sides are parallel, coin- cide in direction. Thus, in the above figure A POA and ';. 42e° y Google EEGnLAB POLYGONS Proposition V. Theorem 432. Two regular polygons of the same number of sides re similar. Given K and K' two regular poJyf>ons, eaoli of n sides. To prove if and K' similar. Proof. Eaoh Z of A'^ -■■ '' '" " -■-- -" ■■- • - -. Art, 174 (i» an equiangular pohjgon ff n sides, each Z = — - — - — ) Similarly each Z of IC— " " ' — Hence K and K' are mutually equiaog'ular. As, i, Art. igs AB_ Also AB = BC .-. fi;,.l. , Art. 417, Ax, ii. And a-b:b'<7 ,. ff;.l. (Why!) . AB^A'B' ^|, AB _ BO "" fiC B'C' A'B' B'C' (Why ?) In like BO CD BE ^ manner ^gJ^^-p^-J^,, elo. Henee K and K' have tlieir homologous sides propor- Hence K and K' are similar. AH. 221. Q. E. D. 433. Cor. The areas of lico regular polygons of the same mmibe.r of sides are to each other as the squares of any two homoloyoHs sides. y Google 2G8 BOOK V. TLANE GEOMETRY Pkopositiox VI. Theokeji 434. I. The perimeters of ttco regular polygons of the same numher of sides are to each other as the radii of their circumscribed circles, or as the radii of their inscribed circles; 11. Their areas are to each other as the squares of these radii. Given AC and A'C two regular polygous, each of n sides, with centers and C, and with perimeters denoted by P and J", radii by E and R', and apothems by r and »■', respectively. To prove. I. P : F' = R -. E' = r -. r'. II. Area AO -. area A'C' = R^ : R'-^r- -. r'-. Proof. I. Thepolygons ACand A'Care similar. Art. 432. Hence P : P-^AB -. A'B'. Art. 34T. But, in the A OAB and (yA'B', I AOB = lA'O'B', [foreai-h Z=l^). Art, 42G. Also OA : OB^O'A' : O'B', (for each A is isosceles). .-. A OAB and O'A'B' are similar. Art. 327. .-. AB : A'B'=OA : CfA'. Art. 321. And AB : A'B'=OL i O'L'. Art. 338. .-. P : F^OA : O'A'^OL : O'L'. Ak. i. Or F: P'^B: R'^r: r'. II. Area AC : area A'C'='AB^ -. A'B'^. Art. 399. But AB^ ■.A^'- = R' : R'" = r^ : r'-. Art. 314. .'. area AG ; area A'C'^B- : £'• ; r : r'^. As, l. y Google EEGULAE POLyGONS ZW Proposition VII. Theorem 435. If the numher of sides of a regular inscribed poly- gon he indefinitely increased, the apothem of the polygon approaches the radius as a limit. Given the regular inscribed polygon AB . . . D oi n sides, with radius OA and apothem OL. To prove that, as n is indeiinituly Ln(;reascd, OL ap- proaches OA as a limit. Proof. In the A OAL, AL> OA — OL, Art. 93. {any siite of a ^ is greater than the lUffcrencn hchrccii the other two sides.) But, asn^^,AB = .: AL^O. Art, 25a, 3. Hence OA—OL = 0. :. OL = OA, or )- = JJ. Oi' the limit of thf apotliera OL is the nidius OA. Q. s. J>. 436. Cor. As« = ^, iJ^— f- = 0. For, iS^ — »-2={K + r) (iJ — r). But, as »(=», B+r^R + E or 2 R, and [i — r = 0. As, s. .-. S2^,-- = 2i," X 01- 0. Ai<. s. Ex. A pjiir of bomologous sidfis of two regiiiai' pentagans are 2 and 3 ft. Find ttn) ratio of tlie areaa of llio yglygons. y Google liOOK V. PLANE GEOMETRY Proposition VIII. Theorem 437. The hnffHi of any line incloning Ihe circrim/crence of II circle, and Jtot passing within the circumference, is greater than tlie length of the circumference. Given the cirenmferenee AKEF, and ABCDEF any line which does not pass within AKEF. To prove cireumferenee AKEF < perimeter ABCDEF. Proof. Let ST be any point on the circumference AKEF not touched by the line ACBEF. At K draw a tangent to the given circle, meeting ACEE at B and I>. Then the straight line BKB < line BCD. Art. 15. To each of these unequals add the line IlEFAB. Then line ABKDEF < line ACBEF. Ax. 9. Henee every, line enveloping the circular area AKEF, except the circumference AKEF, may be shortened. Hence the circamference AKEF is shorter than any line enveloping it. '^ " Q, S. B., y Google REGULAR POLYGONS 271 438. Cor. 1. The circumfef&iice of a circle is less than the perimeter of any polygon circumscribed about the circle. 4.89. Cor, 2. The circumference of a circle is greater than the perimeter of any polygon inscribed in the circle, for each side of an, inscribed polygou is less than the aro subtended by it. 440. Cor. 3, The difference between (he perimeters of an inscribed and a circumscribed polygon is greater than the difference between either perimeter and the circumference of the circle. Proposition IX. Theorem 441. If the number of sides of a regular inscribed, or of a regular circumscribed polygon be indefinitely increased, I. The perimeter of each polygon approaches the circum- ference as a limit; II. The area of each polygon approaches the area of the circle as a limit. Given a circle of eircnmfei-ence (7 and area A, with regu- lar inscribed and cireumscribed polygons, each of n sides, with their periraetei's denoted by P and P', and tijeir areas by AT and K', respectively. To prove that as n is indefinitely increased, P and P' y Google 2i2 BOOK V. PLAXE c.T-,o^iEn;Y each approaches C as a limit,, au,l K and K' each ap- proaches A as a limit. Proof I. Denote the apothems of tlie two polygons by H and )■. Then j,=''.- Art-™- Hence ^^ =^^- '^''- ^^"■ Let 11 = 0=; then Ji—r = 0. Art. 43ii. Rut p'— < f"— 7' (Art. 440) .-. 1"—C^0, or F = C. Also C— P < i"— .?' (Art, 440) .-. C — P=n, or P^C. K Let H =1. :o ; then 7^- - (■= i 0. An. 43G. ... ^^'" "T . ..ll i 0. Art- 253, 4. (/(J)' Wic d(noinin'i(or , J'', infciscs). H^nee — J=-^ i (As S), Z. K' ~ K= 0, (Art. 253, i). But ,ff'— ^1 < iL'— .ff(Ax. 7) .-. .B:'— A=0, or ff' = J.. Al::;o A — K<K'—K{k:^.r) :. A — K=0, or K^A. Q. E. D. Ex, Find the perimeter and firea of a square field, one of whose BldBS is 10 rods. Find the same in a square iield, oni: of wlioso sides is 20 rids. Is it more economical, thei'eforf, to leuee land in largs oc small gelds t y Google REGULAR POLYGONS Proposition X. Theorem 442. Two circumferences have the same ratio as (kei radii, or as thHr dia Given the circles and (7, with eireuniferenees denoted by Caiid C , mdii by R and }}', diameters by D and i/, re- Kpoci.ively. To prove C : C = R : R' = D : D'. Proof. Ill the given ® let regular polygons of the same number of sides be inscribed. Denote the perimeters of the inscribed polygons by P and P', respectively. Then F' It' lience, by alternation, JT^jv,' ■A"''- 30'/. If, now, the number of sides of the similar inscribed poly- gons be increased indefinitely, P becomes a variable ap- proaching C as a limit. Art. 441. P Hence— becomes a variable approaching — as a liri ' E ' Ji" But the v; riable — =the variable C c R ' R' :. c : e- -R: R Also %-,-- VL Art. 315.) (' y Google 2 1 4 ROOK y. I'LANK flKOMiyi'KY rROPOSITlON XI. TnEOIlKM 443. In Cirri/ circle, the ratio of the circumfcrenre to Ikc^. diameter (that is, the nuniber of times the diameter is contained in the circumference) is the same, and may be i by an appropriate symbol {ti). Proof. Given two © with cireumferences denoted by G and C, and diametei's by V and D', respectively. To prove | -|-» C D By alternation Jt^T/ Art- 30T. Hence, in any given circle, — has a vahic equal to that .■. the value of — is the same in all circles. Denoting this constant by ti, in every eircle"^ = 7t. Q. £. B. 444. Formula for the circumference in terms of the ladius. 1) yGoosle EEGULAU POLYGONS 445. Formula for length of arc of a circle. arc _ central Z circiiiufereEoe 360° PiiorosiTioN XII. Thkorkh 446, TJir arra of a regular polyjon is equal to oiie-luilf llhe product of its peritiieter hy its iipoihem. Given the regnhir polygon ABGJ>E with area denoted by K, perimeter by I', and apothem by r. To prove £=JPX?-. Proof. Draw the radii OA, OB, 00, etc., dividing tho polygon into as many A os, the polygon has sides. Ail the ^ thus formed have the same altitude, r. Art.308. .■. the area of each A = i product of its base by r. Art. 3S9. Hence the sum of the areas of the & = | product of the sum of the bases of the A by r, or= % PX-r. But the sum of the areas of the & equals the area of the polygon. Ax. 6, Hence A'='. V X )■. An. s. Q. E, D. 447. Def. Similar sectors are sectors iu different cir- cles which have e{|ual angles at the center. 448. Def. Similar segments are segments in different circles whose area subtend o'liud angles ut the center. y Google TliOrOSITIOX XIII. TUEOKEM 449. T!ir arm nf a circle is equal to one-half the prod- i€t of it-1 circumference hij ils radius. Given a O with fiircunifciTiiee denoted Ity C, radius by E, and avea h_v K. To prove K= i C X B. Proof. Cireiimscvibe a regular polygon about the given eircle, and denote its perimeter by P and its area by K'. In this ease the apothem of the regular polygon is R. Hence E'^iPXK. Art. 44fi. Let the number of sides of the circumscribed polygon be increased indefinitely; then iT' becomes a variable approaching £^ as its limit; Art. 441, Pbeeomes avariable approaching Gas its limit; Art. 441. And viiriiilik'^i'Xfliipproitches I Ox ^iis a limit. Art. 253,2. K' = i i' X A' always. Art 446. But Uenc: --\Cx R. 0. E. I 450. Formula for the area of a circle in terras of the radius li. K^i CXB; but C=27lR. Art. 444. .-. -i(27r-fi)-K. Or K^TiE''. Ax. e. Again E-J i> .-. .fi^^-^-- As. 8. 461. Cor, 1. The area of a circle is equal to the square of the radius, multiplied hy n; or to one-fourth the sgiiare o/ the diameter, muUiplied hy %. y Google MEASUEBMENT OF THE CIECLE 277 452. Cor, 2. The areas of two circles are to each other as the squares of their radii, or us the squares of their diam' eters. K TtIP E- K inJ)'^ D" For 453. The area of a sector is equal to one-half the prod- uct of its radius by its arc. 1 { ceiiti-al I \ 454. Cor. 3. Similar sedor^ squares of their radii. Ex. 1. The measurement of tiie area of a circle can be reduced to the measurement of the length of what single straight line ! Can it he reduced to the measurement of any otiiev single straight line I »nt of a sijigle curved line? Find the area of a circle, Ex. 2, Whose radius is 10 ft. Ex. 6. Whose radius is 2li ft. Ex. 3. Whose diameter is 10 ft. Ex. 7. Wliose radius is E; IB; Ex, 4. ■\Vhose radius is li ft. Et, 8. Wliose radius is Ei/5. Ex. 5. Whose radius is iJ> ft. Ex.9. Whose diameter is *Bi/3. Ex. 10. If the radius of oue cirele is 10 times as gfeat as the ra- dius of another circle, how do their areas compared Also bjw do their circumferouees compare ? Ex. II. A wheel with 6 cogs is geared to awheel with 4S eogs, How many rovolutious will the sniaUer wlieel make while the larger Wheel revolves once ? Ex. 12. A 2 in, pipe will diseharKe how much more water in a ^ven time thau a 1 in, liipij f y Google 2 /« BOOK V. PLANE GEOMETliY Propositiom XIV. Theorem 455, 77»' areas of iito similar serjmeutsare tcreach other as (he squares of their radii. A Given the siraiiar segments ABC and A'B'C in ciroici, 'liose radii are B and R'. To prove segment ABC -. segment A'B'C' = B~ It'"^. Proof. Draw the raGii OA, OB, O'A', O'B'. Then the sectors OAB and (yA'B' are similar. And A OAB and O'A'B' are similar. . sector OAB R' , A 0.4 R I. 448, 447. Alii. 327. t. 454, 338. " scetoc O'A'B' E'-' A O'A'B' . seetor OAB _ A OAB " sector O'A'B' A O'A'ii'' . sector 0-4 B i sector O'A'B' " A O^ii A O'A'B' . sector OAB— A OAJJ__sector O'A'B'— A O'A'B' segment ABO __ segment A'B'C A OAB ~ A 0'.d'B' ;ment ABO _ f A OAB \ ^Kf . meat A'B'C \A O'A'B' J W^ rts. 307,308. Q. £. I). y Google CONSTRUCTION PBOBLBMS CONSTRUCTION PROBLEMS Proposition XV. Problem 456. To inscrihe a square in a given circle. Let the pupil supply a aolutiou. 457. Cor. By Msecting the arcs AC, CB, BD, etc., and drawing chords, a regular octagon may ie inscribed i» the circle; by repeating the process, regular polygons 0/I6, 82, 64, . , . and 2" sides may be inscribed, where n is a positive integer greater than 1. How eau a regular polygon of 2" sides be cireuinscribed aboiit a given circle 1 PROPOiilTION XVI. PROBLEJI inscribe a regular JK^xagon in a given, circle. Let tbe pupil aiipply a Hointion. y Google BOOK V. ri.ASE GEOMETllY of a regular innrrihed hexagon 459. Cor. 1. The si< equals the radius of the c 460. Cor. 2. By joining ihe alternate vertices of a regular inscribed hexagon, an equilateral triangle can he in- scribed in a given circle. 461. Cor. 3. By bisecting the arcs AB, BC, CD, etc., and drawing chords, a regular polygon of 12 sides can be inscribed in a given circle; by repeating the process, regular polygons of 24, 48, ... 3 X 2" sides can be inscribed. Similarly, how can a regular polygon of 3 X 2" sides be circumscribed about a given circle ? Proposition XVII. Problem 462. To inscribe a regular decagon in a given circle. Given the circle 0. To inscribe a regular decagon in the given O. Construction. Draw any radius OA, and divide OA in extreme aud mean ratio at K, OK being tbe greater seg- ment. Art. 371. With A as a center and OK as a radius, describe an arc cutting the given circumference at B. Then the chord AB J5 a side of the decagon required, y Google CONSTRUCTION PROBLEMS 281 Proof. Draw OB and KB. Then OA -. OE^OK : KA, Constr. But AB^OK, .: OA : AB-=AB -. KA. Ax. y. Also lOAB^lKAB. idont. .-. A OAB aud EAB are similar. Art. 327. /. ^0 — ^ABK [homolog. A of similar &). And A AKB is isosceles, Art. 321. [bdiig simHar lo A AOB, ttfticA is isosceles), ;. AB^KB=KO. Ax- 1. Ileiicu 10= IKBO. (Why?) J .-. A0AB=2 10. Art. yti. [ 10^ 10. Adding, Z AiJO + Z OAS + Z 0- 5 Z 0. But Z ABO + Z OAB + Z 0=2 rt Z . Art. 134, .-. 5 Z0=2rt. A. Ax. I. .-. Z 0=i of 2 rt. A , or iV of 4 rt. A . :, are AB is iV of the eiretimf ereneo , Hence, if the chord AB be applied ten times in suc- cession to the circumference, a regular decagon will be inscribed in the given circle. An, 418. Q. E. F. 463. OoR. 1. By joining the alternate vetikes of a regular inscribed decagon, a regular pentagon can be inscribed in a given circle. 464. Cor. 2. By bisecting the arcs AB, BC . . . and draiving chords, a reyular polygon of .20 sides can he in- scribed in a given circle; by repeating iJte process a regular polygon of iO, 80, ... 5 X 2" sides can be inscribed. How can a regular polygon of 5 X 2" sides be cireuia- scribed about a given circle ) y Google 28!^ liOOK V. PLASE OEOMKTHV Pkoposition XVIII. Pkoblej! 465. To inscribe a regular polijgon of fifteen sides (pen- tedecagon) in a given circle. Given the O AK. To inscribe a regular pentedecagon in the given O . Construction. Draw the chord AC equal to the radios of the given O, and the chord AB equal to a side of a regular decagon inscribed in the circle. Art. 462. Draw the chord BC. Then BC is a side of the required pentedeeagon. Proof. .AC is a side of a regiiiar inscribed hexagon. Art. 459. .■, arc AC=i of the circumference. , In like manner are AB = iVof the circumference. Art. 4G2. .", arc BO—t — "i\, or tV of the circumference. As. 3. Hence, if chord BC be applied fifteen times in succes- sion to the circumference, a regular pentedeeagon will be inscribed in the given circle. Art. 418, Q. E. F. 466. Cor. By bisecting the arcs BO, CD, . . . eic, drawing chords, and repeating the process, regular polygons of '60, 60, . , . 15 X 2" sides can M inscribed in a given circle. How can a regular polygon of 15 X 2" sides be circum- scribed about a given circle ! y Google COMPUTATION PROBLEMS 283 COMPUTATION PKOBLEHS Proposition XIX. Problem 467. Given the side and raditis of a regular inscribed polygon, to find the side of a regular inscribed polygon ofdouUe the nuttier of sides, in terms of the given quantities. Given the circle with radius R, AB a side of a regular inscribed polygon, and AC a side of the regular inscribed polygon of double the number of sides. To determine AC in terms of AB and E. Solution. Draw the radii OA and OC. Then OC is the X bisector of AB. (Why () But arc ACB < semieircumferenee. (Why ?) .: AC < a quadrant. Ax. 10. .*. Zj4 00 is an acute Z. Art. 257. Henee,in A 010, Zc^=OP+0C^— 2 OCX 07'. Or AC^^2E^ — 2BXOD But, in the rt. A OAD, 7)& ^OA^—AlP , Art. 347. Or OF = R' — \iABY-. Ax. 8. .: OD^V'E-~i JF^i V'ili^ — AB'. Substituting for OD its value thus obtained, A<f = 2E^ — R-l/i^ —A^. AC-= VKi^E—ViB^ — Aff), yGoosle 284 BOOK V. I'LAXE GEOMETRY 468. Special Formulas. If tho radius B, lie takt>n as 1. AC = \^2 — y4: — AB\ If a side of tlie regular inscribed polygon of n sides be denoted by S„, and ^=1, then S-y,,^^'^ — VT^^Sl Proposition XX. Problem To compute approximately the numerical value ofn. Given a O whose radiiis is 1, and whose eireumferenoe ia denoted by C. To compute C, i. e., 27t, and henee find the miniorieal value of 7t approximatelj-. Computation. 1. Inscribe a regular hexagon in the given circle, and denote its side by &■ Then ySc = l. Art, 459. .', perimeter of the inscribed hexagon = 6. 2. Inscribe a regular polygon of double the number (12) sides. Then, by the second formula of Art. 4GS, Si2=V2- 1/4^^=0.51763809 +. Denoting the perimeter of a regular inscribed polygon of n sides by P„, Pi2 = 12(0.5176380D+) = G.211t!5r08. yGoosle MAXIMA AND MTKIMA S. In the second formula of Art. 468, let h = 12. .-. 834=^2—1/4— (0.51763809+)- = 0.26108238+, etc. Hence Pw = 6.26525772. Computing 8ig, P48, etc., in like manner, the following results are obtained; 5.=l/2- -Vi — i = .51763P09. = .2610523S. = .130S0fi2^. = .0Gr>43817. = .03272346. = .01030228. = .00S18iaj. ■- Fit : P2 •■. F, .". Fa .-. Pi -■■ -Pt ■■• P- = 6,21165708. Su =1/2- -i'4-(.5t76809)' = 6.26525722. ^. =1/2- -V4 — {.2til052;(S)' = 6,27870041. S„ =V'2- -V4 — (.13080(i26)' = 6.28206390. S„.. = VT- -V.i~-{.omi3sny ^ = 6.28290510. -s,„=y'2- -V4 — (.032733-46 J' „ = 6. 28311544. sn>=y'2^ -V 4 — 1.01636228) 6 = S.2S31604I, Bj' continuing the computation it is found that the first six decimal figures in the value of the perimeter of the in- scribed polygon remain unchanged. .-. C, or 271 = 6,283169 approximately. .". 7t=3.14159 approximately, MAXIMA AND MINIMA 470. Def, a maximum (see Art, 268) is the great- est of a group of magnitudes, all of which satisfy certain given conditions, ThuB, the diameter is the maximum oliord wLicli imn be drawn in 471. Dep. a minimum 13 the smallest of a group of magnitudes, all of which satisfy certain given conditions. Thus, of all lines which can he dnira from a given point to ft given Hue the perpendicular is the miniinum. Certain masiiua and minima have already been studied, and we now proceed to investigate more particularly those relating to regular poljgons and tiie circle. y Google Hi BOOK V. IT AXE GEOMETKY Proposition XXI. Theorem 472. Of all trinngles which have two sides equal, tJutt ■i'liK/lr ill which these sides include a right angle is the Given the A ABC and A'B'C in wliicli AB^A'B', OA = C'A', and /.A' is a rt. Z. To prove A A'B'C > A ABG. Proof. In the A ACB, draw CD 1 AB. Then CD < CA. (Why?) .-. CD < CA'. Ax. 8. But the &. ABC and A'B'C have equal bases. Hyp, .■. these ^ are to each other as their altitudes, CD and CA'. Art. 391. .-. A A'B'C > A ABC. (for C'J' > rii). Q. E. B. 473. Iso perimetric figures are figures having equal perimeters. El. 1. Find the iven parallel lines. 1 between two Ex. 2. What is the largest stick that can be placed on a reetangu- lar table 12 I 5 ft,, and not have an end projeuting over a side of the table f y Google MAXIMA. AND MIKIMA 287 Proposition XXII. Theorem 474. Of. all isopfrimetric Mangles which have the same ase, the isoscelea triangle is the viaximitm. I Given the & ABO and AUT) having the same base AB and eqnal perimeters, AC=CB, and AD and DB ub- equal. To prove A ACB > A ADB. Proof. Produce AC to S, making CF=AG. Draw FB. and from Z) as a center, with a radius equal to T>B, describe an are catting FB produced in G. DrawDG and AG. Draw CS and DK 1 AB; also Gi and DP 1 FG. Then Z A£f is a ri?bt Z , Art, 2G1. ( for it maij be inscribed in a semioirHe whose center is C and s ACF). Also ADG is not a straight line, (/or, if it were, the A DAB and DBA WOiM be complements of the = A DGB and DBG, respectwely, and hence icoiikl be equal, aitd .: A DAB would be isosceles, which is contrary to the hypothesis). :. .1F=AC+ CB=AD-\- DB = AD + DQ. Cocstr. Hyp. But AD+ J)G > AG (Art. 02). :. AF > AG. Ax, 8. .-. BF > BG. Art, 111. .-. i r,F > h BG, or Cfl > KD. Ax. 10. .-. A ACB > A ADB. .\yt.:m. (J. E, O. y Google 288 book v. plane gkomktiiy Proposition XSIII. Tueoeem 475. Of hopfirimetrin pohjgonx haviny the same number of Hides, Ihf maximum is eqniUittral. Given ABODE the maximum of all polygons having a given perimeter, and a given number oE sides. To prove -4 BGDB equilateral. Proof. If AfiCDE is not equilateral, at least two of its sides, as AB and BO, must be uneqnal, If this is possible, on the diagonal -iOas a base, con- struct a triangle having the same perimeter as ABC, and having side AB'^B'C. Then A AB'C > A ABC. An. 474. To each of these uneqnals add the polygon ACDE. :. AB'CDE > ABODE. As. a. But this is contrary to the hypothesis that ABODE is the maximum of the class of polygons considered. Hence AB = BO, and ABODE is equilateral. ^___ "!- E. ». Ei. Of all circles which are described on a given liue aa chord, y Google maxima and minima 289 Proposition SSIV. Theorem 4'76. Of all polygons hmnng all sides given hut one, ihe ■maximum can he inscribed in a semicircle having the unde- termined side as a diameter. Given the polygon ABCDEF, (he maxitnnra of all poly- gons having the sides BC, CD, BE, EF, FA, in common, and J.B undetermined. To prove that AB is the diameter of a semieircle in ■which ABCBEF Qs.x\ be inBcribsci. Proof. Bravv lines from any vertex, E, to A and B. The A BEA must be the maximum of ail A having the sides BE and EA, (for, if it is nnt, hj incrmsitig or decreasing the angle BEA, thf A KEA can be chaitgeA till it is a maximum, ihe rest of figure, BCDE a'nl EFA, laeaniime remmniug vnckanged; thus the area of the poly • gon ABCDEF iBould 6e increased, wkich is contrary to Ihe b'jpotliesis that ABCDEF is a raaxiiHum) . .'. Z BEA is a right Z . Art 47^. .". E is on the aemicircumference of whiuh AB is the In like manner the other vertices, C, D and F, must lie on the semicircumference which has AB for a diameter. y Google BOOK V, TLASE GKOMETHY Proposition XXV. Thbokkh 477. Of all polygons formed ; that which can he inscribed in a ri, iih the same given sides, "If in the maxivium. Given ABCDEa. polygon wliich cau be iuacribeii iu a O, and A'B'CD'E' a polygon wliich has the same sides as ABCBE, but which cannot be inscribed iu a O. To prove ABGBE > A'B'G'iyB'. Proof. From any vertex, A, of ABODE draw the diameter AK, and join K to the adjacent vertices C and D, Upon CD' constrnet the triangle C'E'D' equal to A CDK, and draw A'ff . Then area ABGK > area A'B'G'K'. Also area AEDK > area A'E'D'K'. Adding, ABGKVE > A'B'G'E'D'E'. But A CED=A G'K'D'. Subtracting, ABODE > A'B'CD'E'. Art. ^76. (Why t) (Why?) (Why T) Q. E 478. Note. It might happen that one nf the parts of the soeond figure formed by the diameter A'E', aa A'B'C'E', eould be ioEcribed ia a Bemicircle, and .". = ABCK. How, then, mowid the above proof be modified t 479. Cor. Of all isoperimftric polygons of a given niimier of sides the maximum polygon is regular. For it is equilateral (Art. 475), and can be inscribed in a sircle (Art. 477), and is, therefore, rfifiiilar (Art, 417). y Google MAXIMA AND MINIMA lifl Proposition XXVI. Theorem 480. Of two isoperimetric regular pnhjrions, that ivhich has the greater number of sides has (he greaUr area. Given Jta regular polygon of any number of sides, as three, and K' a regular polj'gon of one more, or four sides, and let K and K' have equal perimeters. To prove K' > K. Proof, Prom any vertex, C, of K, ili-fiw a line CD to any point D of the side AB, which meets one of the sides of ZC. Construct the A DCF, having CF^DA, and DF^CA. :. A DCF= A CDA. Art. I01. Adding A CBD, DFCB o K. A^. 2. Henee the polygon DFCB has the same perimeter as K' , and the same area as K. But DFCB is an ii-regukr, while K' is a regular poly- gon of four sides. .-. K' > DFCB. Art, 470. .-. if > A'. Ax. 8. In like manner it may be shown that a reguhir polygon of one more, or five sides, >K, and so on, Q. E. B. 481. Cor. Of isoperiineiric plane figures, (he circle is tile maximum. That is, the area of a circle is greater than the area of any polygon with equal peiimeter. y Google 292 BOOK V. FLAME GEOJIETEY Proposition XXVII. Theorem 482. Of two equivalent regular polygons that wMch h the less number of sides has the greater perimeter. e' , — p — H Given ^and E' two regular polygons having the same area, and having their perimeters denoted by P and P', but £7 having the less number of sides. To prove I" > P. Proof. Let ^ be a regular polygon having the same number of sides as K', and the same perimeter as K. Then K > S. Art. 480. .-. JU > H. Ax. 8. .-. P' > P. Art. 3P9. El. 1. Find the area of a trij and 12 in., and the im^ludifd Siiig triangle having two aides of 6 an Ex. 2. How long is the feni g!e in which two of the sides are 6 is 90". Find the area of another 12 in,, and the inuluded angle 60". about a garden 60x40 ft.? How many square feet in the area of the garden t Find also the length of fence and area of a garden 50 ft. square, Ex. S. Find the area of aa equilateral triangle, a square, a heia= gon, and a circle, in each of which the perimeter is 1 ft. Ex. 4. Find the perimeter ot an eqnilateral triangle, a square, and a circle, in each of which the area is 24 sq, in. Sx. 6. What principle of ma^sima and minima is illustrated in eaob ot the four preceding Exs.? y Google 483. Symmetry of polygons. Many of tlie propertiea of regular figures can be obtained in a aimple and expedi- tious way by the use of the ideas of symmetry. 484. An axis of symmetry is a line snch that, if part of a figure be folded over upon it as an axis, the part folded over will coincide with the remaining part of the figure. EXERCISES. CROUP 46 Ex. 1. How many axes of symmetry liaa an Ei, 2. How many lias a square? a rcgulai' pentagon ! Ex. 3. How many lias a regular liesagoii ? a rtgutar heptagon f Ex. 4. How many haa a regular uctagon ! a regular polygon of n sides T Ex. 5. How many has a circle f 485. A center of symmetry for a polygon is a point such tliiit jiiiv line di'awn throiigli the point and terminated by the perimeter is bisected by the point. squa EXERCISES. CROUP 4 equilateral trlanglH a center r pentagon a centei' ;.f syn letry f Han a iBtry ? Haa a enter of sym- Ex. 2. Has a n regular hexagon ? Ex. 3, 111 general, which regular polygons have a c metry, and which do not ? Ex. 4. Has a circle a center of symmetry t Ex, 6. W'liich is the moat eymiuetiical plans figure atudied thus far f y Google EOOK V. TLANE GIlOMliTiiY SYMMETKY WITH RESPECT TO A LIME OK AXIS 486. Two points symmetrical -with respect to a line or axis are poiuts such tliat the straight Hue joining ^ijem ig bisected by the given liue at right angles. Thus, if PP' is bisected by AB, and PP' is X AB, the poiuts Pand P' are symmetrical with respect to the axis AB. 487. A figure symmetrical with lespect to an axis is a figure such that each point in the one part of the figure has a point in the other part symmetrical to the given point, with respect to an axis. 488. Two figures symmetrical with respect -^f^^^-^-^^c to an axis are two figures such that each point in the one figure has a point in the other figure symmetrical to the given point, with respect to an axis. li' SYMMETRY WITH EESPECT TO A POITJT OR CEKTER 489. Two points symmetrical with respect to f^ a point or ceater are points such that the / straight line joining them is bisected by the / point or center. Thus, if PP is bisected by the point (7, C is / a center of synunetry, with respect to P and P. -y 490. A figure symmetrical with respect to a point or center is a figure such that each point in the figure has another point in the figure symmetrical to the given point with respect to the center. 49 1 . Two figures symmetrical with respect to a center are figures such that each point in one figure has a point in the other figure symmetrical to it with respect to the center. -^'^i?^^ y Google Proposition XXVIII. Theorem 492. If a figure is symmetrical with respect to tivo axes tcMch are perpendicular to each other, it is symmetrical with respect to their point of intersectimi as a center. Given the figure ABO . . .H symmetrical -with respeot to the two axes XX' and YY'; and XX' 1. YY' and intersect- ing it at 0. To prove ABC ... if symmetrical with respect to 0. Proof. Take any point P in the perimeter of the figure, and determine the points F' and P", symmetrical with respect to P, by drawing FKP' L XX', and PLP" ± YY' . Draw KL, P'O. OP". Thon Pr [1 i'i", and PP" || XX'. But PK^KP. Also" Pfi:=and!i LO. :. i'P = and || LO. :. KLOP is a Cn , andA'i = and || PO. In like manner it may be shown that KL = B.ih .: PO=nnd II OP". Ax. Art. 121. Art 48fi. Art. 157. A X. 1. Art 160. 1 01 Art 122. Hence any straight line drawn throngh 0, and terminated by the perimeter, is bisected at 0, . is a center of symmetry for the given figure. Ar!. 4Eio, y Google I'LANE GEOMKTili" EXERCISES. CRO Ex. 1. The diagonals uf a regular pentagon Ex. 2. 1( ABODE is a regular penta' .rmnjrit.. ■lUK a Ex.3. Ill the same figure Bt' = irC. Ex. 4. Also KCDE is a parallblograiii El. 5. Also JC=JB + t!A-. Ex. 6. Tlie diagonals of a regular p( Ex. 7. The apotliei lugle equals one-half of a Ex. 8. Tho altitude of au equilateral ti equals one and a half times the radius of t eumserjbed circle. Ex. 10. The side of an equilateral triangle equals Ry"i, tl oribed circle being denoted by It. El. 11. Theapothom hiilf the side of a r^uiar , regular inscribed hexagon rihed triangle { = -x]/'^)- Ex. 12. The side of a reg side of the regular inscribed triaogle. Ex. 13, Find the side of an insc the side of a circumscribed square. 1 of a terms of E. 13 of the ins rihed square in tarn I inscribed square, i ribed and circumscribed squares one of these is double the other. a of a regular inscribed triangle ia '-'-j — ' y Google ESEKCISES. THEOREMS Ivi Ex. 17. Shovf that the area of a regular inscribed hexagon is •^ — . What, then, ia the ratio of the area of a regular inaoribed tri- angle to that of a regular iusi^ribed hexagon 1 Ex. 18. The area of a regular inscribed hexagon is three-fourtha the area of the regular circumscribed hexagon. Ex. 19. The area of a regular inscribed hexagon ia a mean pro- portional between the areas of a regular inseribed and a regular cir- cumsoribed triangle. LSuo. On a figure similar to that of Prop. IV, p. 2(i6. let OP inter- seet AB in K, and compare the A OKA, GAP, and OPA' .'[ "Ex. 20. The area of a regular inseribed polygon of 2n sides is a mean proportional between the areas of regular inscribed and ciruum- seribed polygons of n sides. Ex. 21. The area of a regular inscribed oetagoa equals the area of the rentaugle whose base and altitude are the sides of the oircum scribed and inscribed squares respectively. Ex. 23. An angle of a regular polygon is the supplement of the angle at the center. Ex. 24. Diagonals drawn froni a vertex of a regular polygon of n sides divide the angle at that vertex into ?i^2 equal parts. Ex. 25. The diagonals formed by joining the alternate vertices of a regular hexagon form another regular hexagon. Find also the ratio of the areas of the two heiagons. Ex, 26. If squares be erected on the sides of a regular hessgon, the lines joining their exterior vertices form a regular dodecagon. Find also the area of this dodecagon in terms of 6, asideof the hexagon, Ex. 27. The square of a side of an inscribed e^juilateral triangle equals the square of a side of an inscribed square added to the square of a side of an inscribed regular hexagon. y Google NR CiEOMETliV Ex. SO, Given AaBhC, AcB, BdC, somi- eirnles; prove that the sum of the two cres- cents AeBa and BdCb equals the ateaoE the right triangle ABC, Ex. 31. An equiangular poly)?oii in- scribed ia a oircla is regular if the number o£ it [SUQ. In the figure to Prop, III, p. 264, t :. are AE^&sc BO, .". aide Jif^side BC, ete,] sides be odd. ^ JEUC = ai-e EDOB EXERCISES. GROUP M JlASl.MA A^^D MINIMA Ex. 2. Of equivalent piirallelogra Ex, 3, Of i 30 peri metric ri Ex. 4, Divide a giveu lir ■s whicli 13 t Ex, 5. Find a point in the hypotenuse oi the 8um of the squares of the perpendieular the legs shall he a minimum. Ek. 6. How shall a mile of wire fence t triangle such thnt 3 from the point to Ex. 7- Find the area in acres included by a mile of wire (enae if it be stretched as a square, a regular hexagon, and a circle respectively. Ex, 8, Of all triangles with the same base and equal altitudes, the isosoelcs triangle has the least perimeter, Ex. 9, Of all polygons of a given number ot aidta inscribed in a given circle, the maximum is regular. [SuLi. Prove the maximum polygon (1) equilateral, (3) equi- angulaj,] y Google EXEIJCTSES. SVMMETKY 299 E3. 10. Find tlie masimiuii leetatigle inscribea in a circle. Ex. 11. Find the maximum rectaugle that eaii bo inscribed iu a [Sue. Inscribe a square in tbe circle.] Ex. 12. Of trapsKoida inscribed in a semicircle (liaviug the diam- ter us one base), find the maximum. Ex. 13. Divide a given straight line into two pnrts such that the am of tiie aquarea ol these parts sliall be a minimum. EXERCISES. CROUP SO SYMMETRY Ex. 1. A rhombus has how many axes oi! symmetry? Has it a center of symmetry ? Ex. 2. "What asls o£ symmetry has a quadrilateral which has two pairs of equal adjacent sides f Has such a figure a center of symmetry ? A parallelogram is symmetrical with respect to Ihe point .1 — of its diagonals. Ex. 4. A segment of a circle is symmetrical with respect to wha xis! Ex. 5. Has a trapezium a center of symmetry ! An axis o Ex. 6. How many axes of symmetry have two equal circles take as one figure 1 Have they a center of symmetry I Ex. 7. "What axis of symmetry have any t*o circles f Ex. 8. How must two equilateral triangles be placed so as to hav & center of symmetry f So as to have an axis of symmetry f Ex. 9. It two polygons are symmetrical with reference lo a een ter, any two homologous sides are equal and [>arallel and drawn i opposite direetioua. y Google lOK V. I'LANE GEOMETUi EXERCISES. GROUP 81 VARIOUS THEOREMS if a eirtle equals 'wice tlie an inscribod equilateral triangle la to the Ex. 3. Tho diagoaal joining any two opposite voL'litts of a regulaf Ex. 4. Tha radius of an inaeribed regulsL' polygon is a mean pro- portional between its apotbem and the radius of tlie uireumacribad regular polygon of the same number of sides. Ex. 5. If the sides o£ a regular hexagon be produced, their points of intersection are the Tortices of another regular hexagon. Also Had the ratio of the areas of the two hexagons. Ex. 6. Of all lines drawn through a given point within an angle, and terminated by the sides of the angle, the line which is bisected at the given point cuts off the minimum area. Ex. 7. Eaeh angle of a regular polygon of n + 2 sides contains _ rjgiit angles. Ex. 8. The diagonals from a vertex of a regular polygon of ii -[- 3 sides divide the angle at that vertex into « equal parts. Ex. 9. The sum of the perpendiculars drawn to the sides of a regular polygon of n sides from any point within tho polygon e(|uals n times the apothem. Ei. 10. An equiangular polygon circumscribed about a circle is [SUQ. Draw radii from the points of contact, and lines from the vertices of the polygon to the center.] Ex. 11. An equilateral polygon circumscribed about a polygon is regular if the number of its sides is odd. [Sua. See Figure of Prop- III, p. 264. Prove Jr=iiy, Draw radii and prove /.T= IQ, etc.] y Google EXEECISES. MISCELLANEOUS THEOREMS 301 Ex. 12, How must two equal isosceles triangles be placed so aa to have a center of symmetry f How masl they be placed so as to have an axiS of symmotry ! If, in a regular inscribed polygon of n sides, •%, denotes a side and r™ denotes the apothem, show thjil. Ex. 13. For the triangle, S3 = Rv''3, r,, = i5. Es. 14. For the square, S, = Bv'2, i-. = iJtv'3. Ex. 15. For the heiagoa, S6 = K, r6 = iBv'3- Ei. 18. For the octagon, Ss~RV'2—i/2. Ex. 19. For the dodecagon, Su =• EV'2—t/Z. Ex.30. Prove that V = K' + Sio'. Ex.21. Trove that Si,' = S„' + S|f,'. Ex. 22. IE ADB, AaC, CbB are semicir- cles and DC ± AB, prove that the area bounded by the three semicircumferencea equals the area deaoribed on I>C as a diameter. ■^ ij Js Ex. 23. If p„ denotes the perimeter of an iaseribed polygon of n sides and P„ the perimeter of a oitonmaoribed polygon of n sideB, 1p^ P^ y Google 302 BOOK V. I'LAKE GEOJIETIii EXERCISES. CnOUP S2 PROBLEMS Circumapribe about a ftivon eire5e Ex. 1, An equilateral triangle. Ex. 2. A square. Ex. 3. A regular p':^ntagon. Ex. 4. A regular hesagou. Ek. 6. A regular dpcagon. Ex. 7. Constpuet a regular pentagram, or .i-poinled star, Ex. 8. Conatruet a he.Yagram, or O-pointed star. Ex. 9. Constniet an S-pointed star. Ex. 10. Construct iin angle of 36". Ex. 11. Construct angles of 18', 9", 72°. Ex. 12. Construct angles o£ '2i'', 12°, G", 48°. eireumferenpe into two parts which shall Ex. 14. Construct a regular pentagon which shall have twice the perimeter of a given regular pentagon. Ex. 16. Conatruet a regular pentagon whose perimeter shall equal the sum of the parlmeters oC two given regular pentagons. Ex. 16. CocBtruot a regular pentagon whose area sliail be twice the area of a given regular pentagon, Ex. 17. Construct a regular pentagon whose area shall be equal to the sum of the areas of two given regular pentagons. Ex. 18. Constru oamferenee of a giv y Google EXERCISES. PROBLEMS 303 Ex. 19. Construct a circle whose area shall be three timeH the area of a given circle. Ex. 20. Construpt a ciroumterenee equivalent ti the =iim and another equivalout to the diBereneo, of two gi\en i irtmnleiencoa ] five equal parts by Ex. 21. Construe a circle e equivalent to th e difference, o£ tv Ex. 22. Construe a circle wh ofagi e„ irole Ex. 23. Bis ct t!i B iiroa of a ference. Ex. 24. Divide th e area of a drawin geo ncen rice rcnmferene Onag ven line cons ruct Ex 25. Ar gular pentagon. Ex 26. Ar egula hexagon. Ex 27. Ai egula dodeeagon Ex 28. Ac rcle qiiivaleiit t Ex 29 Ins criLe a regular oo Ex 30 lus cribe a circle in a Ex 31 Ins eribe a square in mieircle. ;iven square. Ex. 32. In a given equilateral triangle inscribe throe equal circlsa each of which touches the other two circles and a side of the triangle Ex. 33. In a given circle inscribe three equal circles which shal touoh eaeb other and the given ci renin ferenoe. y Google NUMERICAL APPTJCATTOXf! OF PLANE OEOMETEY METHODS OF KUMEEICAL COMPUTATIONS 493. Cancellation. In numerical work in geometry, aa elsewhere, the labor of computations may frefjuentlj- be economized. Those methods of abbreviating work, which are particularly applicable in the ordinary numerical appli- cations of geometry, may be briefly indicated, as follows: To simplify numerical work by caneeliation, group together as a whole all the numerical processes of a given proilem, and make all possible cancellations before proceed- ing to a final nuvterical redtiction. Ex. YinA the ratio of tlie area of a rectangle, whose base and alti- tude are 42 and 24 inches, to tlie area of a trapezoid, whoae bases are 21 and 35 and altitude 12. Bj Arts. 383, 394 J area of i-eetanRle ^ >2X 24 _ ^ ^fk?* ^r, j^,. area of trapezoid 6(214-35) ^X^B 494. Use of radicals and of 71. Where radicals enter in the course of the solution of a numerical problem, it fre- quently saves labor not to extract the root of the radical till the final answer is to be obtained. Ei. 1. Find the area of a circle oiroumacribed about a squara whose side is 8. The diagonal of the square must be 8i/2 (Art. 34C). .-. the radius of O =4i/2. .'. by Art. 449, area of G = x(4i/2)' = 32ir = 100.6. Area. y Google NUMERICAL COMPUTATIONS 305 Similarly in the use of tt, it frequently saves labor not to substitute its numerical value for n tilt late in the process of solution. Ei. 2. Fiud the radius of a circle whose acefi is equal to the suia of the areas of two circles whose radii are 6 and 8 inches, respectively, Deaote the radius of the required eirele by x. Then, by Art, 440, ^ j' = 36 tt + C4 ,r. .-. Tl'= 100 TT. ,-. j' = 100, and x= 10, Fadiu.^. 495. Use of x, j, etc., as symbols for unknown quan- tities. In some cases a numerical computation is greatly facilifated by the use of a specific sijwhol for an imknoirti quantity. Ex. In a triangle whose aldea are 12, IS, and 2.'), find the segmeots of the side 25 made by the bisector of the angle opposite. Denote the required segments of Bide ^5 by Then 12 : 18 = j: : 2;') — j:(Art. 332) .-. 18j- = 12(2.j-j.-) (Art. 302) .■.:r = 10.i And 25~a: = lf..)" ' r Sc(jmfiits . 496. Limitations of numerical computations. Owing to the limitations of human eyesight and of the instruments used in making measurements, no measurement can be accurate beyond the fifth or sixth figure; and in ordinary work, stteh as is done by a carpenter, measurempnts are not accurate beyond the third figure. As all numerical applications of geometry are based on practical measure- ments, it is not necessary to carry arithmetical work beyond the fifth or sixth significant digit. Other methods of facilitating numerical computations, as by the use of logarithms, are beyond the scope of this book. y Google 6Ub PLANE GEOMEIKY KUMEBICAL PKOPEKTIES OF LINES EXERCISES. CflOUP S3 THK KlfiHT THIANGLU Ek. 1. Find Uie hypotenuse of a riglit lria»i;lo wht.so leE3 ave 13 and :i5. Eji. 2. The hypotenuse of a right triangle i? 29. and one leg ia 20. Find the other leg, Ei. 3. If a window ia 15 ft. from the (■round and the foot of a ladder is to be 8 feet from the house, how long a. ladder is necessary to reaeh the window P Ex. 4. Find the diagonals of a rectangle whose sides are .j and 13. Ex. 6. Find the diaconnl of a square whose side is I ft, 6 in, Ex. 7. The diagonals of a rhombus .ire '2i and 10. Find a side, Ex. 8. One side of a rhombua is 17 and one diagonal is 30. Find tiie other diagonal. Ex. 9. In a circle whose radiua is 5, find the length of the longest and shortest chords through a point at a distance 3 from the center. Ex. 10, In a circle whose radius is 25 in,, find the distance from the center to a chord 48 in. long. i in. fio Ex. 12. A ladder 40 ft. long reaches a window 20 ft. high on one Bide of a street and, if turned on its foot, reaches a window 30 It, high on the other side. How wide ia tlie street J Ex. 13. If one leg of a right triangle ia 10 aud the hypotenuse is twit;e the other leg, hnd the hypotenuse. Ex. 14. Find the altiludo of an equiiaterai triangle whose side isG, Ex. 15, Find the aide of an equilateraltriaugle whose altitude is 3, y Google NUMESICAL EXERCISER. LINES .307 Ex. 16. Find the side of & square whose diagonal ia 15. Es. 17. One leg of a right triangle is 3, and the sum of tha hypotenuBs and the other leg is 9. Find the sides. Ex, 18. A tree SO ft. high ia broken off 40 ft. from the ground. How tar from the foot o£ the tree will the top strilce V Ex. 19, The radii of two oirolea are I and 6 in., and their oenters are 13 in. apart. Find the length of the common esternal tangent. Ex. 20. TiiG Bides of a triangle are 10, 11, 12. Find the length of the projection of the side whose length is 10, on tlie side 12. EXERCISES. CROUP S4 TRIANOJ-ES IN GENERAL Ex, 1. The sides of a triangle are 13, 18 and 20. Find the aeg- fflents of the side 20, made by the bisBOtor of the angle opposite. Es. 2. In the same triangle, find the segments of the Bide 20, made by the bisector of the exterior angle opposite. Ex. S. If the legs of a right triangle are C and 8, find the hypote- nuse, the altitude on the hypotenuse, and the projections of the legs on the hypotenuse. Ex. 4. Is a triangle acute, obtuse, or right, i£ the three sides are 5. 12, U; 5, 11, 12; 5, 12, 13; 4, 5, 6? i are G, T and 8, compute the length Et. 6. Also the length of the median on the same side. Ex. 7. Also thu length of tlie bisector oE the angle opposite the iiJe B. Ex 8. It two Bides and a diagonal of a paiulloiotsram ftre 8, 13 ^ia 10, find the other diagonal. [SuG, Uao Art. 352.] y Google PLAKE GEOMETRY El. 11, The hypoteniisB of a right triangle 'in 10, and the altitude on the hjpotenuaa is 4. Find tho aegments of the hypotenuse and tlie legs. El. 12. Find the three medians, the fliree bisectora, and the tLreu altitudes ol a triangle whose sides are 13, 14, 15, EXERCISES. CROUP SS CIRCTTMFRRENCF.S AM> ARCS T. - 55 Lamg fl-^.-yt Ex. i. Find thn ciroumEerence ot a eirclf whose radius in I ft. 9 in. Ex 2. Find the radiuB of a circle whose Rireumferenee is 121 ft. Ex. 3. A biojele wheel 28 in. in diameter makes, in an afternoon, 3,000 revolutions. How far does the bicycle travel F Ex. 4. What is the diameter of a wheel whkh makes 1,400 revolu- tions in going B,800 yds. ? Ez. 5. If the diameter of a circle is 20, find the length of an arc of 60°; also of 83°. Ex. 6. If the length of an arc is 14 and the radiua is G, End the namber of degrees in the arc. Ex. 7. If the arc of a quadraut m 1 ft. in length, find the diameter. Bk. 8, Two eonoentrie circumferences are 88 and 132 in. in length, respectively. Find the width of the Bircular ring between them. Ex. 9. If the year be taken aa 365i da., and the earth's orbit a circle whose radiua is 93,250,000 miles, find the velocity of the earth in its orbit per second. y Google NUMERICAL EXERCISES. LINES 309 Find the r&dius and circumference of a circle circumscribed about Ex, 10. A square whose aide is 5. Ex. 11. An equilateral triangle nhose side is 4. Ex. 12. A rectangle whose sides are 12 and 5, Ex. 14. Find the diameter of a cirt ;le eircH angle whose sides aro 7, 15, and 20, Ex. 15. Find the radius of a oirclf i whose the perimeter ol a square whose diagoni il is lU. EXERCISES. CROUP liS CBOKDS, TANGENTS, AND SECANTS ES. 1. Two intersecting chords of a cirule are 11 and 14 in., and the segments of the first chord are 8 and 3 in. Find the segments of the second chord. [St;G. Denote the required segments bf :c and 14 — e,] Ex. 2. In a circle whose radius is 12 in., a chord 16 In. long is passed through a point 9 in, from the center. Find the segtnents of the chord. Ex. S. Two secants drawn from a point to a circle are 24 and 37 in. long. If the external segment of the first is G in., find the external segment of the second. Ex. 4. From a given point a seeant whoso external and internal segments are 9 and 16 is drawn to a circle. Find the length of the tangent drawn from the same point to the circle. Ex. 5. Frenj a given point a tangent 24 in. long is drawn to a circle whose radius ia 18 in. Find the distance of the point from the center. Ex. 6. I( a diameter GO in. long is divided into 5 equal parts by «l»ords perpendicular to it, find the lungth of tiu) chowlB, y Google PLANE GEOMETKY Ex. 8. If the earlh is will tbe light of a HgLlhou EXERCISES. CnOtIP Bt LINES IN' SLMILAli FICURiiS Ex. 1. K the Hides of a triiwgle are fi, 7 and S, and the shortest Bide of a similar triangle is 18, find the other aidea of tbe secoLd Ex. 2. If a post 5 ft. high easts a shadow :! ft. lotij;, find the height of a ateeple wkiph easts a shadow 90 I't. long. Ex. 3. In a triangle whose base is 14 and aliitude 1!!, a line is drawD parallel to the base and at a distauee 3 froni the base. Find tbe length of tbe line thus drawn. Ex. 4. The upper and lower bases of a trapezoid are 13 and % and tbe altitude ia H. If the legs are produced till they meet, find the altitnde of eaph of tbe two trianglts thus formed. El. 5. If the upper .-ind lower bases of a trapezoid are i, and li, and the altitude ia h, find the altitude of each of tbe triangles formed by producing the legs. Ex. 7. If the perimeter of a legular poljgon is three times the the ratio of their apoihema t Es. 8. It the eireumterenees of two eirelea are 600 and 400 ft., wbat is the ratio of their diameters 1 Ex, 9. In the preceding example, if a ohord of the first circle ia 30, what is the length of a chord in the second circle, subtending the same number of degrees of arc 1 y Google NUMERICAL EXEECISES, AREAS 6 COMPUTATIOW OF AREAS EXERCISES. CROUP iiS AREAS OF TRIANGLES i of a triangular field whose b El. 2. Find the area oC a triangle wlioae aldea are 10, 17, and El. 3. Find the area in aerts ot a liuld whose sides are (iO, ud no chains. d of a triaag-ulur tield t^xch i>S «li clialcs. Find the area of Ex. 5. Ad isosaelea triangle whose base \3 16, and eai^li of wh legs la 34. Ex. 6. Aa equilateral triangle whoEe altitude is 8. Ei. 7. A right triangle in which the segments of tha hjpotec made by tlie altitude upon it are 12 and 3; also, in one in which Ex. 8. An isosceles right triangle whose hj-potennse is 12. Ex. 9. A right triangle in which fhe hypotenuse is 41 and one leg is 9. Ex. 10. Find in two ways ibe area of a triangle whose sides are 6, 5, 5. Ex. II. A side of a given equilateral triangle is 4 ft. longer than the altitude. Find the area of the triangle. Ex. 12. The area of an isosceles triangle is 144 and a leg is 24. Find the base. Ex. 13. The area of an equilateral triangle is 4t/3. Find a side. Ex. 14. The area ot a triani;ie is 1125, and if.h: f = 2 : ;i : 4. Find Ex. 15. The area o£ a triangle is G sq. in., and two ot its aides are il aud 5 in. Find the remaining eida. y Google i'LANE GEOMETHY EXERCISES. GROUP SB AREAS OP OTHER RECTILINEAIE t ■ir.URES PimJ tlie area of Ex. 1. A par«.lleloi;ram whosa Ijaso is 2i ft. fi ' tuaei8l2ft. Oiu. ill. and w Ex. 2. A trapt^zoia whose uasvs are 12 ami 20 tude is n ft. in. and w Ex. 3. A vhombus whoae diagonals are 9 ft. and 2 yds. Ex. 4. A quaJrilateral in wliieh the sides All, BC, CD, DA a 13, 14, 15 and the diagoual AC is 17. Ei. 5. A quadrilateral iu which the sides ai-o £7, 36, 30, 2 the angle included betweeu the first two siciea is a right angle. Eit. 6. A square whose diagonal is 12 in. Ex. 7. Piud the number of boards, ea<;li 4 yds. long and wide, which are neeessarj to cover a floor 48 X 24 ft. Ex. 10. A rectangular garden contains 4,524 aq. yds. and 13 20 yds. longer than wide. Find its dimensions. Ex. 1 1. Each Bide of a rhombus is 24 ft. and each of the larger angles 13 double a smaller one. Find the area. Ex. 12. Find the area of a rhombus one of whose sides is 17, and one oC whosB diagonalB Is 3U. Ex. 13, The area of a trapeaoid is 4 acres, one base Is J2U yds., and the altitude is 100 yds. Find the other base. Ex. 14. The bases of an iaoseeles trapei^oid are 20 and 30 and the legs an IT. Find the area. y Google NUMERICAL EXEKCISES. AREAS 313 Ex. 15. The base of a triangle !a 20 and tbe altitude 13. Find the length of a line parallel to the base ivhieh cute off a trapezoid vi\ioRe area is 80 sq. £t, [Suo, Denote the altitude ot trapezoid by 18— a; and Gad its •■ppBr base by similar triangles.] Ex, 16. The perimeter of a polygon, circumscribed about a circle trhose radius is 20, is 340, Find the area of the polygon. Ex. 17. The area of a rectangle is 144 and the base is three times the altitude. Find the dimensions. Ex. 18- Find theareaof aregularhesagon oneof whose sidesialO. Ex. 19. Find the area of a regular deeagou inscribed in a eircie whose radius is '20. Ex. 20. f iud a aide of a. regular hesagoii whose area is 200 sq, in. EXERCISES. CROU? 60 AREAS OF CIRCULAR FIGURES Ex. 1. Find the area in aeres of a circle whose radius is 100 yds, Ex. 2. Find the radius in inches o! a circle whose area is 1 sq. yd. Ex. 3. Find the area ot a circle whose circumference is j). Ex. 5. Find the radius of a circle whose area equals ^e areas of three circles whose radii are 20, 2H, 10. Ex. 6. In a eircie of radius 50 find tbe area of a sector Ex. 8, IQ a circle whose radius is 7, the area of a sector is 43 sq. ft. Find the number of degrees in its angle. Kx. 9. Ill a circle whoso radius is 10, fiud the sum ot the segmeiitB formed by au inseribed square. y Google PLANE GEOMETHV Ex. li. The same pond is surrounded by a driveway 30 ft. wide. Find the area of the driveway. Ex. 12. Two tangents to a nirele, whose radius is 15, include an angle of 60°. Find the waa Included between the tangents and the radii to the points of contact. Ex. 13. Find the length of the tether by which a cow must be tied, in order to graze over eiiactly one acre. Ex, 14, Three equal eii'cles touuh each other externally. Show that the area included between them iaJtHyS — ^)- Kx, 15, If the area inclnded between three equal circles which touch each other externally is a square toot, find tlie radius of each EXERCISES. CROUP ei AREAS OF SIMll.AK FIGURES Ex. 1. The homoloKons sides of two similar triangles are 3 and 5. Find the ratio of their areas. Ex, 2. The liomologoua sides of two similar polygons are 4 and 7, and the area of the first polygon is 112, Find tlie area of the second polygon, Ex.3. The radius of a circle is e. Find rlie radins ot a circle hav- ing three times the area of the given circle. Ex, 4. The areas of two circles are as IC to p, and the radius of the first is 8. Find the radiua of the second. Ex. 6. The sides of a triangle are 5, G, 7. Find the aides of a similar triangle containing 9 times the area of the given triangle. Ex. 6. If, ia finding the area of a circle, a student uaeB 1> = 50 as K = 50, how will the area as computed differ from the correct area 1 y Google MISCELLANEOUS NUMERICAL EXERCISES Et. 7. In a triangte whose base \a 24 in. and altitude is 18 ii ftltituiie is bisected by a line purallel to the base. Find tbe a the triangle cut off. Ex. 9, In a circle wboBe diameter is 30 in,, what are the diameters of ooncentric eireumferences which divide the area into three equiva- lent parts T Ex. 10. If a circle be eoostrueted on the ridius o! a given circle, Hud segments, one in each eirGle, be formed by a line drawn from the point ol contact, find the ratio of the , EXERCISES. CROUP 63 OENERAL NUMERICAL EXERCISES EH PLANE OEOMETRY iiid the base is IQ. e equivalent to this triangle. Ex. 4. The Rides of a triangle arc 7, 8 and S inches. Find the aides of a triangle of four times the area. Also, of twice the area. Ex. 5, The temple o! Herod is said to have aooommodated 210,000 people at one time. If each person required 27 X IS in., and one-third the space iiiaide the temple be allowed for walls, sanctuaries, etc., what were the dimensions of the temple, if it was a square f mgle are 12, 16 and 21., vrhat are the / the bisector of the angle opposite t lateral in order m aiif;le of (10°, Find the diameter o£ a wheel which, in a mile, maires 480 y Google PLASE GEOMETKV Find the radius of a circle equivalent to Ex. 10. A square mlioae side is 10. Ei. II, An eijiiilateral triangle whose siile Is 12. Ex. 12. A trapezoid whose bases aro IG and 18 and altitude 9. Ex. 13. A aemicirelo whoso raijins is 15, '!■ of ft circle whose area shall be e<]uiTa- I whose diameters are U4 and 17. Ek. 15. A oirele, a square, and an equilateral triangle each, have s perimeter of 12 yds. Find the area of eai^h figurii. Ex. 16. In a circle whose area is 400, the area of a seetor is 125. Find the angle of the sector. Ex. 17. How many acres are included w track, if the traek is in the shape of a rei it IE wide 1 Ex. 19. Find the area of the circle circumscribed about a angle whose sides are 40 and 9, Ex. 20. One leg of a right triangle teen the hypotenuse and the other leg Ex. 21. Find the area of an isosceles Ex. 22. In a triangle whose sides are 16, 18, 20, find the length of the altitude, median, and bisector of the angle opposite the longest Ex. 23. A line 16 inches long is divided internally in the ratio of 3:5; find the segments. Also find the segments when the line is divided externally in the same ratio. Ex. 24. A line 16 Inches long is divided in extreme and mea& ratio. Find the segments. y Google MISCELLANEOUS NOMEEICAI, EXERCISES 317 Ex, 25. If a line is divided in extreme and mean ratio and the Bmaller segment ia 4, flnd tlio whole line. Ex. 28. In a circle whose diameter is 20, a chord is passed throuKh a point at a distance 6 from the center, perpendicular to the diameter through that point. Find the length ot this chord, and ot the chords drawn from its extremities to the ends of the diameter. En. 29. Eiioh Itg of an isosceles trapezoid is Jll, and onH base exceeds the other by 16. Find the altitude. Ex, 30. It three arcs, each of 60° and having 10 for a radius, are to the other two, find the area included by them. Ex. 31. Find the area of a trap? izoid whose legs i ire 4 and 5, and whose bases are 8 and 11. Ex. 32. A square piece o£ land : and a<-irpnlar pi, ice each contain 1 acre, llow many more feet of fenc does one require . than the other ? El. 33. IE the base of a triangle is doubled and the altitude re- mains unchanged, how is the area affected ! If the altitude is doubled and the base remains unchanged t It both the base and the altitude are doubled f ot Ex. 37. Find the side of an equilateral triangle equivalent to a oitcle whose diameter is 10. Ex. 38. The area of a rhombus ia loti sq. in., and one side is 1 ft. 1 in. Find the diagnnuls. y Google 318 PLAN'E GEOMETRY Ex. 39. Tlie EJiipB of atriansle are S, in, 12, Find the aicis of the triangles iii.icle by llie bisector of the angle opposite Ihu side 12. Ex. 40. In a circle of area 275 sq. ft., a re<^tangie of area 150 aq. ft. is inaeribed. Sliow how to find the sides of the reetaogle. EXERCI3E3. CROUP 63 EXERCISKS INVOLVING . THE METRIC SYSTEM Ex. 1, Find the area of a triangle of which the baee is 16 dm. and the altitude SO cui. Ex. 2. Find the area of a triangle whofe sides are 6 m., TO dm., 800 cm. Ex. 3. Find the area in square meters of a circle whose radius is 14 dm, Ex. 4. If the hypotenuse of a right triangle is 17 dm, and one log iB 150 cm., find the other leg and the area. Ex. 5. It the circumference of a circle \i 1 m,, find tlio area of the circle iji square deeimeters. Ex. 6. Find the area in heetares, and also in acres, of a circle whose radius is 100 m, Ex. 7, If the diagonal of a rectangle is 35 dm, and one side is 800 mm., find the area in square meters, and also in square inches. Ex, 8, Find the area of a trapezoid wiiose bases are 600 em. and 2 m., and whose altitude is SO dm. Ex, 10. In a given circle two chorda, whose lengths are 15 dm. and 13 dm,, intersect. 3f the segments of the first ehord are 12 dm. and 3 dm., find the segments of the second chord, Ex. 11. Find in det imetera the radius of a circie equivalent to a square wbosa side is 1 ft. 6 in. Ex. 12. Find in feet the diameter of a wheel which, in going 10 kilometers, makes 5,000 revolutions. y Google SOLID GEOMETRY Book VI LINES, PLANES AND ANGLES IN SPACE DEFimTIONS AND FIRST PRINCIPLES 497. Solid Geometry treats of the properties of space of three dimensions. Many of the properties of space of three dimensiona are determined by use of the plane and of the pcopertiea of plaue figures already obtained in Plane Geometry. 498. A plane is a surface such that, if any two points in it be joined by a straight line, the line lies wholly in the surface. 499. A plane is determined by given points or lines, if no other plane can pass through the given points or lines "without coinciding with the given plane. 500. Fundamental property of a plane in apace. A plane is determined by any three points not in a straight line. For, if through a line con- ,c necting two given poiuts, A and B, a plane be passed, the plane, if rotated, can pass through a third given point, C, in but one position. ■'* The importance of the above principle is seen from the faet that it reduces an unlimited surface to three points, thus making a vast econoijij to the attention. It also enables us to eouuect different planes, and treat of their properties aystematically. (3iy) yGoosle 320 BOOK VI. SOLID GEOMETRY 501 . Other modes of determining a plane. A plane may also be delerm'uied by any equivalent of three points not iu a straight line, as by a straight line and a point outside the, line; or hy two intersecting straight lines; or by two parallel straight lines. It is often more eonvenient to use one o£ these latter raethods of determining a plane than to reduea the data to tliree points md use 502. Representation of a plane in geometric figures. In reasoning concevuiug the plane, it is often an advantage to have the plane represented in all directions. Hence, in drawing a geometric fignre, a plane is usually represented to the eye by a small parallelogram. This ia virtually a donble aae of two intersecting lines, or of two parallel lines, to determine a plane (Art. 501]. 503. Postulate of Solid Geometry. The principle of Art. 499 may also be stated as a postulate, thus: Through any three points not in a straight line ior their eqnivaJeni) a plane may be passed. 504. The foot of a line is the point in which the line intersects a given plane. 505. A straight line perpendicular to a plane is a line perpendicular to every line in the plane drawn through its foot. A straight line perpendicular to a plane is sometimes called a normal y Google LINES AND PLANES 321 506. A parallel straight line aud plane are a Hue aad plane which cannot meet, however far they he produced. 507. Parallel planes are planes wliich caonot meet, however far they he produced, 508. Properties of planes inferred immediately. 1. A atraighl line, not in u yircti plane, can in!erseci IJie given plane in but one point. For, if the line intersect the given plane in two or more points, by definition of a plane, the line must lie in the plane. Art. 498. 2. The intersection of two planes is a straight line. For, if two points common to the two planes be joined by a straight line, this line lies in each plane (Art. 498) ; and no other point can be common to the two planes, for, throagh a straight line and a point outside of it only one plane can be passed. Art. soi. Ex, 1, Give an esample o£ a plane Burfa<ie; of a eiu'vod siirfaee ; of a Burfaee, part plane and part curved; of a surface composed of different plane Burfaees. Ex. 2, Four points, not all in the same plane determine how many different places ? how many different straight lines ? Ex. S. Three parallel straight lines, not in the same plans, deter- mine how many diHerent planes ? Ex. 4. Four parallel straight lines can determine how many difter- Ex, 5. Two iiilerseetins straight lines an.! a point, not in thei!,' plane, determim; how many difie-rent planes 1 y Google 322 BOOK VI. tiorjD geometky Propositio>- I. Theokem 509. If a siraiglht line is perpendicular io each of ttvs other straight lines at their point of intersection, it is per- pendicular to the plane of those lines. Given AB J. lines BC and BD, and the plane UN pass- ing through BC and BD. To prove AB ± plane MN. Proof. Through B draw BG, any other line in the plane MX. Draw any couveuieut line CD interseeting BC, BG and BB in the points C, G and B, respectively. Produce the line AB to F, making BF^^AB. Connect the points C, G, D with A, and also with F. Then, in the A ACD and FCD, CD=CD. Ident. AC- CF, and AD^BF. Art. 112. .-. AACB^AFCB. (Whyf) .-. / ACB= I FCD. (Why!) Then, in the A ACG and FCG, G6=CG, (WbyT) AC^CF, and / ACG = IFCG. (Why!) .-. A AGG^AFCG. (Whyt) yGoosle LINES AND PLANES 323 .-. AO=GF. (Why?) . B and 6 are each equidistant from the points A and F. :. BO is X AF; that is, AB ± BG. Art. ii3. .-. AB X plane jl/iV, Art. 505. (forii is X any line, BG, in the plane MJS, through its fool). q. E. D. PROrOSITION IT. TlIEOItKM 510. AU tiie perpeitdieulars that can be tirawn to a given line at a given point in the line lie in a plane perpen- dicular to the line at tJte given point. X ■<ri\. Given the plane M]V and the line 5(7 both X line JBa,i the point B. To prove that BG lies in the plane MI^. Proof. Pass a plane AF through the intersectinff lines AB and BC. Art. 503. This plane will intersect the plane MN in a straight line BF. Ari. 508, 2. Bat AB X plane MN (Hjp.) .". AB X BF. Aft. 505. Also AB X BC. Hyp. .-. in the plane AF, BC and BF X AB at B. /. BG and BF coincide. Art. 71. But BF is in the plane MN. .: BC must be in the plane 31N, {for B(J uMWiklcs wUh BF, ichidi. ties m the jilaiic MX). e, E. B. yGoosle 824 BOOK VT. SOTJD GFXl^ilETliY 611. CoE. 1, At a given point Bin iliestm'ijhl line AB, io construct a plane, perpendicular to the line AB. Pass a plane A-F through AB in any convenient diveiiliou, and iq the plane AF at the point B coiiBtruet BF ± AB (Art. 274). Pass another plane through AB, and in it construct BP X AB. Through the lines BF and BP pass the plane MX (Art. 503). MX is the f required plane (Art, 509). 512. Cor. 2. Through a given external point, P, to pnsa a plane perpendicular to a given line, AB. Pass a plane through AB and P (Art. 503) , and in this plane draw PB ± AB (Art. 273) . Pass another plane through A B, as AF, and in AF draw BF X AB at B (Art. 274). Pass a plane through BP and BF (Art. 503). This will he the plane required (Art. 509). 513, Cor. 3. Through a given point hut one plane can be passed perpendicular to a given line. Ex. 1. Five points, no tour of which are in the same plaoe, deter- line how many different planes ! how many different straight lines 1 A atvaight line and two points, not le, determine how macy difiecent plat Ex. S. In tte figure GDF are equal. y Google LINES AND PLANES Proposition III. Problem 514. At a given point m a plane, to erect a perpendicu- lar to the plane. 1 ''• 1 sz^^^ \-^v i\ Given the point A in piano MN. To construct a line pei-pendicular to MN at the point A. Construction. Through the point A draw any line CD in the plane MF. Also through the point A pass the plane PQ X CD (Art. 511), intersecting the plane J/jV iu the lineiSS. Art, 508, 2. In the plane FQ draw AK 1 line ES at A. Art. 274. Then AK is thi; ± requirud. Proof. CD ± plane Py. Coiistr. ;. CD 1 ,4/t:. Art. 505. Henw AK X CD. But il/r X 7i\S\ Coustr. .-. AK X phine MN'. Art. 5oy. Q. E. F. 615. CoK. .(l( fl given point in a plane but one perpen- dicular to the plane can he drawn. For, if two X could be drawn at tlie given point, a plane conld be passed through them intersecting the given plane. Then the two X would be in the new plane and X to the same line (the line of intersection of llie two planes, Art. 505); which is im- possible (Art. 71). y Google ■rlh BOOK VI. fiOLID CEOJIETilY Proposition IV. Problem 516. Fimn a given point iviiliout a jilatie, to draic a line pcrpfndicifhii- to the plane. Given the plane MX and the point A external to it. To construct from A a line ± plane MX. Construction. In the plane MX draw any convenient iine BG. Pass a plane through BC and A (Art. 503), and 111 this plane draw AD ± BG. Art. 273. In the plane MX draw LD X BC. Art. 274. Pass a plane through AD and LD (Art. 503), and in that plane draw AL A. LD. Art. 273, Then AL is the ± required. Proof. Take any point C in BG except X>, and draw LO uud A G. Then &. ADG, ADD and LDG are right A . Consti .-. J6''=Zd^ + DG'. Art. 400. .-. AC^ = ZL^ + LD' -f 'DC'. Art. 400, Ax. 8. .'. AC^^AL^ -{■ LG'^. Art, 400, As. 8. .-. ZAiCisaright /. . Art. 351, But AL ± LD. Conatr. .-. AL X MX. Art 517. Cor. Bitt one perpendicular can be dratcn from a given external point ia a given plane, y Google LINES AND PLANES 327 Proposition V. Theorem 518. I. Oblique lines drawn from a point !o a plane, meeting the plane at equal distances from the foot of the perpendicular, are equal; II. Of ftco oblique lines drawn from a point to a plane, but meeting the plane at unequal dlniances from the fool of the perpendicular, the more remote is the greater. Given AB ± plane MN, BD=BC, and BH > EC. To prove AD^AC, and AM > AC. Proof. I. In the right A ABD and ABO, AB = AB, and BD = BC. (Why?) .-. AABZ>=A^£0. (Why') .-. AI> = AC. (Wby?) II. On Offtake iSJP^Be, and draw AF. Then AF^AC (lypart of theorrm jin^t prm'f,d) . But AIT > AF. (Why?) .-. AU > AC. As. 8. Q. E. D. 519. Cob. 1. Conversely: Equal oblique lines drawn from a point to a plane meet the plane at equal distances from the foot of the perpendicular drutni from the same point to the plane; and, of two unequal li}iex so drawn , the greater line meets the plane at the greater distance from the foot of the 'perpendicular. y Google 328 SOLID r.EOJlETl'.Y 520. r'dJi 2. The locus of a point in space cijuklhiant from all llif points in the circumference of a circle is it ulraight line passing through the center of the circle and perpendicnhtr to its plane. 521. Cor. 3. The perpendicular is the shorlpst line Ihut can be drawn from a gii-en point to a given plane. 522. ]>KF. The distance from a point tu a plane is the perpendleiihir drawn from the point to the jilniit;. Proposition VI. Theorem 523. If from the foot of a perpendicular to a plane a line be drau-n at right angles to any li7te in the plane, the line drawn from thtpoint of intersection so formed to any point in the perpendicular, is perpendicular to the line of the plane. Given A7i ± plane MX. and BF X CJ), nny Hue in 3LY. To prove AF J_ CJJ. Proof. On €1) take FP and FQ equal segments. Draw A P, HP, AQ, BQ. Then BP^BQ. Art. 112. Hence AP^AQ. Art. ma. .'. io the line AF, the point A is equidistant from P and Q, and F is equidistant from P and Q. .: AF ± €T). (Why!) Q. E. ». Ejc, In the above figure, it Ali^G, AF^S, and jy = lU, find QF, DF and C^. y Google LIKES AND I'LASES Proposition VII. Theorem 524. Two straight lines perpetnUcitlar to ike same plane >■€ parallel. i \ / \ •f / / """"T I Given the lines All ami CD ± plane ilW. To prove AB 11 CD. Proo(. Draw BI), anil Uu-ousli J), in the plane MS, draw FR X BI>. Draw Ai). Then ;j/< J- FII. Con.lr Ai> ± FM. Art. 023. 0/' i F7Z. Art. 505. .-. BB, AD and (7/") rn-p all ± J'J/ at the poiut T>. .". R/', A7' and Cf.) all lie in the same plane. Art. 5io. .-. AB imd CD ave iu the same plan'^. (Why?) But Ali and (JIf are ± B'li. Art. 50o. .-. AB and 6'Z> are 1|, Art. vi\. t). E. D. 525. Cor. 1. If one of tiro parallel liiie^ j.v pn-pendicn- iar to a picnic, the other is perpendicular to the plane also. For. if AB and CI) be ||, and AB ± plane PQ, a line drawn from C L TQ 4 f must be II AB. Art- ;i?4. But CP iiuist coincide with tin? line V ^ ;)\ so drawn (Art. 47, 3); .'. CD 1 7'(^ ^ —''^^ yGoosle :^,:";o SOLID GEOMKTP.V 526. Cor. 2. Ifin-o straight lines are each parallel to a third straight Uhp, tJiey are parallel to each other. For, if a plane be drawn 1 tu llii? 1;liird line, each of the two other Sines must be 1. to it {Art. 525), and therefore be 1| to each other (Art. 5-4). y*iioi'OfiTio\ YIII, Theorem 527. If a sirairjhi line ex/erna! to a given plane is paral- lel lo a line in the plane, then ihejirst line is parallel to the given plane. B Cn in the plane MN. !■ .¥_V. Given tlio straight linpyl/; II To prove Ml \\ plai Proof. Pass a plane thron^h the || lines AB and CD. If AB meets MN it must meet it in the line CT). But AB and CD cannot meet, for they are ||. Art, 120, .". AB and MN cannot meet and are parallel. Art. 506, Q. E. 9. 528. Cor. 1. // a straight line is parallel to a plane, the intersection of the plane with any plane passing through the given line is parallel to the given line. y Google LINES AND PLANER 529. Cor. 2. Through a given line ^_ ,g (CD) to pass a plane parallel to another q given line (AB). q..^i^^,^. r Through P, any point in CD, draw ^ QR |[ AB (Art. 279). Througli CDaui QB pass a plane (Art. 503). Tliis will be tlie plane reqaired (Art, 527). Jt AB and CD are uot parallel, but one pluiie can be drawn through CD\\AB. Proposition IX. Theorem 530. Two planes perpendicular to the sanu re parallel. Given the planes MN and PQ X line A 11. To prove MN \] PQ. Proof. If MN and PQ a.ve not parallel, on being pro- duced they will meet. We shall then have two planes drawn from a point per- pendicular to a given line, which is impossible. Art. 5i3. .■, MN and PQ are parallel. Art. 507. y Google f;oLin (";En:MET];v Proposition X. Thk(jr!:m 631. // two parallel planes are cut by a tJih-d plane, ike intersections are parallel lines. Given M'N' and PQ two II planes iiitersecteil by the planuj BS in tbe lines AB and CD. To prove AB \\ CI). Proof. AB and CT) lie in the same plane US. Also AB and CI) panKOt meet; for if they did meet the planes ilO' and FQ would meet, which is impossible. Art- r>07. .-. AB and ('/> are pm-illoi. Art. 4L 0- E. D. 532. Cor. 1. Parallel lines included between parallel planes are equal. For, if AC and BD are two parallel lines, a plane may be passed through them (Art. 503), intersect- ing MN and BQ in the 11 lines AB and CD. Art. 531. .". ABDC is a pariillelogram. Art. U7. .-. .1(7= CD, Art. 155. yGoosle LISES ASD PLANES 333 533. Cor. 2. Two parallel planes are everywhere equi- distani . For lines 1 to oae of them are il (Art. 524). Hence the segments of these lines included between the || planes are equal (Art. 532). Proposition XI. Theorem 534. If two infersecting lines are each parallel to a given plane, the plane of these lines is parallel to tJie given plane. \ : > 1^ \. '\ ,.■-'1 "■J \ Given the lines AB and Cl>. eatih \\ plane PQ, and inter- secting in the point F: and lilN a plane through AB and CD, To prove MX \\ PQ. Proof. From the point F draw FS X PQ. Pass a plane through FC and Fff, intersecting PQ in BK; also pass a plane through FB and FE, intersecting PQ in EL, Then EK \\ FC, &nd BLW FB. Art. 538. But FE 1 EK and EL. Ait. 505, .-. FE 1 FC ;.iid FB. Art. 123. .-. FE ± MX. Art. 509. .-. MN WPQ. Art, ,^30. yGoosle d'ii BOOK VI. SOLID GEOJIEiny Proposition XII. Theoresi 535. A straight line perpendicular to one of two parallel planes is perpendicular to the other also. \ -- L_4 - \j \. .^••J -1 \ Given the plane MN || plane PQ, and AB L FQ. To prove AB L MN. Proof. Througrb AB pass a plane intersecting PQ and MN in the lines BC and AF, respectively; also throngh AB pass another plane intersecting PQ and MN in BJ) and AH, respectively. Then BC 11 AF. and BD \\ AS. But AB X BC and BB. :. AB 1 ^J'and^fl'. .-. AB X plane J/A". 586. COK. 1. Through a given point to pass a plane parallel to a given plane. Let the pupil supply the construction. 537. Cor. 2, Through a ginen point but one plane can be passed parallel to a given plane. Art 531. Art 505. Art 123. Art 5CB. 0. S. >. y Google LIMES AND PLANEiS 335 Proposition XIII. Theorem 588. If two angles not m the same plane have their corresponding sides parallel and extending in the same direc' Hon, tht angles are equal and their planes are parallel. Y Given the Z.BAC m the plane MX, and the ZB'A'G' in the plane PQ; AB and A'B' \\ iind csteuding in the same direction; and AC and A'C W and extending in the same direction. To prove Z^^iO = Z B'A'C, and plane ilfiV |1 plane P^. Proof. Take AB = A'B\ and AC=A'C'. Draw AA', BB', CC, BC, B'C. Then ABB' A' is a ZI7 , Art. 160, (foi-ABandA'B'are^and [\). :. BB' and AA' are = and ll. Art, 155. In like manner CG' and AA' are = and !|. .-. BB' and CC are^and ||. (Why ?) .-. BGC'B' is a C7 , and BC=B'C'. {Why V) .-. A ABC= A A'B'C. (Why 5) .-. ^A = ZA', (Why?) Also AB II A'B', ,■. AB \\ plane PQ. Art. 527. Similarly AC 11 plane P(?. .-. plane MN || plane PQ. Art. 534. y Google doO BOOK V!, HOLID GEOMKTKY Propositiux XIV. Theorem 539. // two straight Ihies are hilers€cletJ bf/ tlin^r purah lei planes, the corresyoiuVuuj svijairnia »/ tlic-'in lines are proportional. \ Given the straight iities AB and CD intersected by the 1! planes MN, PQ and SS in the points A, F, B, and 0. H, D, respectively. AF CR To prove — : the plane PQ in 0. FB MB Proof. Draw the line AB Interseetiu Draw i^G. BB, GH, AO. Then FG \\ BB, B.\id. GR\\ AG. Art, 531, .■^'=^. Art. 317. FB GD And ^=f|. (W.,n •■ FB SB ' y'l Q. E. B. Ex. 1. la above figure, if AF=1, FB = 5, and CH^S, lind CD. Ex 2- If Cif=3, SD=4, and JB=10, find JJ-' and fif. y Google DIHEDRAL ANGLES DIHEDRAL ANGLES 540. A dihedral aagie is the opening between two in- tersecting planes. Prom certain points of view, a dilitdriii angle may be regarded as a wedge or slice of space cut out by the planes forming the dihedral angle. 541. The faces of a dihedral angle aro the planes form- ing the diliedral angle. The edge of a dihedral angle is the straight line in which the faces intersect. p 542. Naming dihedral an- gles. A dihedral angle may bo named, or denoted, by naming its edge, as the dihedral angle -1J5; or by naming four points, two on the edge and one on each face, those on the edge coming between the points on the faces, as F^AB-Q. The latter method is necessary in naming two or more dihedral angles which have a common edge. 543. Equal dihedral angles are diliedral angles which <ian be made to coincide, 644, Adjacent dihedral angles are dihedral angles liav- ing the same edge and a face between them in common. 545. Vertical dihedral angles are two dihedral angles having the same edge, and the faces of one the prolonga- tions of the faces in the other. 548. A right dihedj'al angle is one of two c((iial adja- cent dihedral iuiglcs t'onued by two ])lanea. y Google 338 vy. 547. A plane per- pendicular to a givoii plane is a plaiii; fovra- ^____„ ing: a right diliedra! \ angle with the givuu \ plane . \ Many of the properties of dihedral angles are obtained most conveniently by using a plane angle to represent the dihedral angle. 548. The plane angle of a dihedral an- gle is the angle formed by two lines drawn one in each face, perpendicular to the edge at the same point. Thus, in the dihedral angle C-AB-F, if FQ is a line in the face AD perpendicu- lar to the edge AB at P, and PR is a line in face AF perpendicular to the edge AB at F, the angle QPIi is tiie plane angle of the dihedral angle C-AB-F. 549. Property of plane angles of a dihedral angle. The magtiltmU of tlio plane angle of a dihedral angle is the same at every point of the edge. For let EAG he the plane I of the dihedral I E-AB-I) at the point A. Then PR || AE, and PQ || AC (Art. ll^l.) .-. IRl'Q - Z.EAC (Art. 538). 550. The projection of a point upon a plane is the foot of a perpendicular drawn from tlie point to the plane. 551. The projectioa of a line upon a plane is the locus of the pro- m jections of all the points of the line on the plane. Thus A'B' is the projection oi AB aa. the plane 31N. y Google DIHF.DIUL AXGLES 60)} Proposition XV. Tiieoreji 552. Tico dihedral angles are equal if their plane angles are equal. lU .LI I Given Z DBF t! . | Iiik- I of the dihedral Z ChAB-F, IB'B'F' the plaiiL. ^ uf the dihedral ^C'-A'B'-F, aud ZDBF==Z.I)'B'F'. To prove Z C-AB-F = I C'-A'Ii'-F. Proof. Apply the dihedral IC'-A'B'-F to IC-AB-F so that Z li'Ii'F coincides with its equal, IDBF, Gaom. As. 2. Then line A'B' mirst coincide with ,4.ii, Art. 515. {fof A'B' and AB are both X jAu.eiJBFut ihepoiiiIJi). Hence the plane A'B'D' will coiueido with plane ABD, Art. 501. (ihroiigh tao taUmecting lines uiih/ one pliiiie mit he [tti^seil). Also the plane A'H'P will coincide with the plane ABF, .". Z (7'-A'/?"-7'"" coincides with Z 6'--4/>'-i'' and is equal to it. Avt. tr. 558. Cor. The vertical dihedral oni/I''^ Jhnu'-d h 11 tin) intersecting planes are equal. In like manner, many otlicr [n'opertii^i^ of phuie aiitrlis are true of dihedral angles. y Google 340 )!noK VI. SOLID 'ieoi]i:jrv Proposition XYL Theorkm 554. Tu-o dihedral aiuih^ li'iv' thr .s^imc r .■-■,^B Given Uie Jihe.lral A V-All-D rikI C'-AT,'-!)' liaviug the plane A fMZ) and C'A'T*', respectively. To prove I C'-A'B'-n' -. lC-Ali-1) =IC'A'!Y -. I CAB. Case I. When the plane A <"A'J>' ami CAT) (Pigs. 2 andl), are comtneiisurahk'. Proof. Find a eommou measure of tlie A V'A'D' aiid CAD, as Z GAK, and let it be contained in Z G'A'D' n times, and in LGAT) m times. Then / C'A'iy : I CAD^n -. m. Through A'B' and the lines of division oE Z C'A'D' pass planes, and through Ali and the lines of division of Z CAD pass planes. These planes will divide the dihedral IC'-A'B>-D' into n, and ZC-AIi-D into m parts, all eqnal. Art, 5S2. .-. ZC'-A'B'-D' : lC-AB-I)=n -. y,i. Hence Z C'-xVB'-D' : Z C-AB-7J = Z C'A'D' -. Z CAD. (Whj ?) yGoosle DlHF-DliAL ANGLES J41 Case II. When (he plane angles C'A'D' and CAB (Figs. 3 and 1) are incommensurable. Proof. Divide the -Z CAD into any miraber of equal parts, and apply one of these parts to the I C'A'D' . It will be eontiiined a i;ertain nnmber of times with a remain- der, as ^LA'D', less than the unit of measnre. Hence the i G'A'L and GAD are commensurable. .-. I C- A'B' ~r,:l C-AB-D^ / fJ'A'L : Z CAD. Case I- If now we let tho unit of measure be iudefiiiitulv iliniiii- iahed, the Zi.-l'/)', which is less tliau the unit of measure, wiil be indefinitely diminished. .-. I CA'L i I C'A'D' as a limit, and iC'-A'ir-L^ lC'-A'B'~D' ■i^s-^WmM. An. 2&L -A'B'-L Hence Z C'A'B'-D' IC-AB-D becomes a variable, vith limit. But the \ tnes a variable with ^C'-A'B'- L _ Z.C- Mi-B I C'-A'B'-B' Ex. I. How many straifjiit lines nre neeessafy to indicate a dihe- dral nngle (as IE-AB~II, p. 338)? IIow many straight lines are neeesBary to indieate the plaoe angle of a dihedral angle 1 Hence, ■what is the advantage of naing a plane angle of a dihedral aogle instead of the dihedral anf;le itfielE f Ex 2, ogoug to p: B thre iial propert; [ dihedral angles anal- y Google Proposition XVII. Theorem 555. If a tstrhUjkt line is pn-pendic\dar to a plane, every plane ilmini fliroiigh iluil line in pcrpemUcular to the plane. Glvea the line Ali X plane MN, and the plane PQ through A B and interseethig MN in EQ. To prove PQ X MX. Proof. Ill the plane .'l/.Vdi-aw BO ± RQ at B. Bnt AB X JiQ. Art. 505. .-. ^ABC U the plane / of the dihedral ZP-RQ-M. Art. 548. But /.ABC is. a visht I , Art. 505. (for All 1 MNhy li-ip.). 556. Cor. A plane perpendicular f" the ed<je of a dihedral angle is perpendicular to eacli of the two faces form- ing the dihedral aixfjle. y Google dihedral anglbs 343 Proposition XVIII. Theorem 557. If two planes are perpendicular to each oilier, a straight line drawn in one of them perpendicular to their line of intersection is perpendicular to Ike other plane. Given the plane PQ X plane 3fy and intersecting it in the line RQ; and AB a line in PQ X HQ. To prove AB X plane MN. Proof. In the plane US' draw BC X EQ. :. ZABCis the plane I of the dihedral ZP-RQ-M. Art. 548. .-. Z^ECisart. /. . Art. 554. {/hi- I'-EQ-Mu a ri'jht <lihe<lna I ). .■, AB X BO and fi(^ at their intersection. .-. AB X plane WN. (Why %) Q. £. 9. 558, Cor. 1, If two planes are perpendictiiar to each other, a perpendicular to one of them at any point of their intersection will lie in the other plane. For, in the above figure, a X erected at tlie point B in the plane MX must coincide with AB lying In the plane PQ and X MN, for at a given point in a plane only one X can 1)6 drawn to that plane (Art. Slij). 559. CcR. 2. If two planes are perpendicular to each other, a perpendicular to one "plane, from a point in the other plane, will lie in the ollu-r plane. y Google DOOK Vf. ^OLID GEOMl^TRY PKOPOSITION XIX, THEOJiEM 560. // iu'O inUrsecHntf plni a third plane, iheh- Unr of ■in the third plane. ', each pBrppnd'cnUir ion is iKvpi-'udicKtar Given the planes PQ and BS 1 plane Mlf, and inter- secting in the line AB. To prove AB ± plane MX. Proof. At the point B in wliich the three planes meet erect a ± to the plane MX. This ± must He m the plane PQ. and also in the plane RS. ^'^^- ^'^'^■ Hence this L must coincide with AB, the iiitersfction of PQ and RS. Art. 50S, 2. .-. AB ± plane .1/-V. g. E. B. 561. Cos. If two planen, indmVnig a right dihfdra! angle, are each perpendicular to a third plane, the inferspc- tion of any two of the planes is perpendicular to the third plane, and each of the three lines of intersection is perpen- dicular to the other lu'o. Ex. 1. X»me all iha dihaiiral an^ Ez. 2. If Z CBQ^SO", find the ri a on the above fijjiire. o£ ea.eli pair of diliedral d. y Google DIHEDRAL ANGLES Proposition XX. Theorem 562. Every point in the plane which iiseets a given dihedral angle is equidistant from the faces of the dikfdral angU- Given plane CB bisectins? tlie diluidi-al I A-BR-T), P any point in phme ISC, I'Q iiud FT 1. faces BA and IW, respectively. To prove I'Q^PT. Proof. Through PQ and FT pass a plane interseetmg AB in QE, BD in KT.and BG in Pli. Then plane FQT J. planes AB and BD. Art. 555. .■, plane PQT ± line RB, the intersection of the planes AB and BD. Art. 5fi0. .-. RB L RQ, EP and BT. Art. r.or.. .-. i QBP and P/C7' ni^e the plane A of tht dihedral A A-BR-P and F-BE-D. But these dihedi-al ii are orinal. .-. LQRV= IPRT. :. rt.A P()/^ = rt, A PRT. :. PQ^FT. Art. 54S. Hyp. 563. n^ tonisof all points of a (lilu'dral angle is the plane t (Why?) Q. E. ». ■qni<li.slant from- Ihe fares ^r:ciiitg ihv ilihi'iirul angle. y Google 1U10K VI. SOLEII XXI. PiiOBLGM 564. Through unij Ktrdi'jht I'li". not perpendicular to < giten plane, to pass a plane pcriKiidkular to the given plane Given the line AB not ± plane MN. To construct a plane passing through AE and ± MN. Construction. From a point -1 In ths liue AB draw a AC to the plane MN. Throngh the intersecting lines AB and AG plane AD. Then AB is the plane required. Proof. The plane AD passes throngh AB. Also plane AD ± plane MN, {M sAC, ichichig 1 MN). Art 51(1. pass tlie Art 503. Conefr. Art 555. Q. E. r. 565. Cor. 1. Through a struigJd line not perpendicular to a given plane only one plane can be pmsed perpendicu- lar to thai platie. For, if two planes could be passed through AB ± plane My, this intersection AB would be X MN (Art. 560), which is contrary to the hypothesis. 566. COE. 2. The projection upon a plane of a straigM line not perpendicular to-that plane is a straight line. For, if a plane be passed through the given line X to the given plane, the foot of a X from any point in the line to the given plane will be in the intersection of the two planes (Art. 5oy), y Google DIHEDRAL ANGLES 347 PROPOsiTiON XXII. Theorem 587. The acute ant/le which a line maki's ivith Us pro- jeetion on a plane is the least angle which it makes with aiiy line of the plane through its foot. Given line AB meeting tlie pianp MN in the point B, ■BC the projection of AB on MN, and PJi any other line ia the plane MN through B. To prove that ZIBC is less than ZAIiP. Proof. Lay off PB equal to CB, and draw AG and AP. Then, in the A ABC and ABP, AB=AB. BG^ BF. Ent AG < AP. .: Z^BC is less than lABP (Why t) Art. 108. Q. E.D. 568. Def. The inclination of a ihie to a plane is the 'ciite angle which the tjiven line makes with its pi'ojsetion 'Pon the given jjlaiie. E». 1. A plane has an inelination of 47" to ea«tt o£ the ta^ne of a iliadral angle and is parallel to clie edge of tbe dilieiiral angle; how aonj degrees are in the pkne UHglt of the dihedral an^'l..'; Ex. 2. In llie fiL-iire ™ pae? 3-ir,. if PT=QT,liow large lathe '''"dral z A -!:il-li'i i[ n---llT, hmv large Is it? y Google SOT.ID gi:u.mk-i'i;y Proposition XXIII. Problem 569. To <lruw a common pTpeu'/'Cidar to any tivo Hi, tiof in the same plane. Given the lines AB and VJJ not in the same plane.' To construct a line perpenLliculiir to botli AB and CD. Construction. Tlirongh AB i>i(..3=! a. plane MXW line CIk Art. j>! Through CD pi^s a piano CF 1 ])laiie M^' (Art. 564), aui interaecting plane MN in the line BF. Then EF \\ CD (Art. 528), .-. EF must intersect Ai (which is not |1 CD by hyp.) in some point K. At E in the plane CF draw LK ± EF. Art. 274 Then LK is the perpendicular reqitired. Proof. LK ± EF. Con^ir .-. LK ± CD. Art. l:.;. Also LK ± piano MX. Art. ss: .-. LK _L line _l7i. (Wl,y^ .-. LK ± both C« and AB. Q. E. ?. 570. Only one perpendicular can be draicn between ttco lines not in the same plane. For, if possible, in the above figure let another Hue BD be drawn ± AB and CD. Then, if a line be drawn throagh B II CD, BD X this line (Art. t2;t), and .-. X plane MN (Art. 509). Draw DF X Hue EF; then Z>F X plane MS (Art. 557), Henoe from the point T) two X, DB and DF, are drawn to the plane MX, which is imposBiblu (Art. 517), y Google rOLYIIEDUAL Af^dLK POLYHEDRAL ANGLES 571. A polyhedral angle is the amount o£ opening between tlireu or more planes meeting at a point. Such an angle may be regarded as a portion of spBoe cut out by the planes torming the anfila. 572. The vertex of a polj-hedral angle is the point in which tlie planes forming the angle meet ; the edges are the lines in which the pianes intersect; the faces are the portions of the planes forming the polj"lie(ival angle which are inclnJed between tlie edges ; tlie face angles are the angles Eormeil by the edget;. Each two adjacent fae.es of a polyliedral angle form a (iihedral angle. The parts of a polyhedral angle ;ire its faee angles ami liihedval angles taken togetlier. 573. Naming a polyhedral angle. A polyhedral angle ia named eitlier by namiuij the vertex, as I': or by naming the vertex and a point on each edge, as V-ABG. In case two or more polyhedral angles have the same Vertex, the latter method is necessary. In the above polyhedral angle, the vertex is Y; the edges are VA, YB. T"C; the face angles are J.rjS, BVC, AVC. 574. A convex polyhedral angle is ii polyhedral angle in which a suction ttiade by a plane cutting all the edges IS a convex polygon, as y-ABCI>E . 575. A trihedral angle is a polyhe- iral angle having tliree t'aces; a tetrahedral angle is one laving four face*, etc. y Google GEOJIETliY 576. A trihedral angle is rectangular, birectangular, or trirectaogular, aeeordinK as it contains one, two, or /ht-ee right dihedral angles. 577. An isosceles trihedral angle i two of whose faee angles are equal. trihcdi-al angle 578. Vertical polyhedral angles an having the same vertex and the faees ( the other produced, 579. Two equal polyhe- dral angles are polyhedral an- gies bavin;^ their correspond- ing parts equal and arranged in the same order, as V-ABC and V~A'B'C. Two equal polyhedral angles may be made to coincide. 580. Two symmetrical poly- hedral angles are polyhedral an- gles having their corresponding parts equal but arranged in reverse order. If the faces of & trihedral angle, V-ABC, produced, they will form a vertical trihedral angle, V-A'B'C, which is symmetrical to V-ABC. For, if V-A'B'C be rotated forward about a horir.ontal aiis through V, the two trihedral angles ai to have their corresponding parts equal I ranged in reverse order. Similarly, any two verlieal polyhedral are aymmetrieai. polyhedral angles : one the faces of y Google P0I.YI!E1>R.\L ANGLES 351 581. Equivalence of symmetrical polyhedral angles. It has been shown in Plane Geometiy (Art. 488) that two triangles (or polygons) symmetrical with respect to an axis have their corresponding parts equal and arranged in reverse order. By sliding two such figures about in a plane they cannot be made to coincide, but by lifting one of them up from the plane- in which it lies and tnrnii be made to coincide with the other figure. Symmetrical polyhedral angles, however, cannot be made to coincide in any way; henee some indirect method of showing their equivalence is necessary. See Ex. 29, p. 358. and Arts. 789-792. it mav 8 figUl Prop. XX. rf those OQ tlie figure ! Kr, 1. Name the trihedral angles i^PBQ^m", what kind of trihedral a If iLPBQ=m°, whot kind are they? Ei. 2. Are two trireot angular trih Prove this. Ex. 3. Are two lines which are perpendicular to the same plane necesaarily parallel f Are two planes which are pei'pendicular to the same plane necessarily parallel? Are two planes which are perpen- dicular to the same line necessarily parallel f Ek. 4. Let tlie pupil out out three pieces of pasteboard of the form ndioated in the aoeoropauying figures; cut them half through where 'ho lines are dotted; fold them and fasten the edges so as to form hree trihedral angles, two of which (Figs. 1 and 2) shall be equal iEd two (Figs. 1 ;ind 3) sjmmetrical. hy esperiment, let the pupil ind which pair may be made to ooinuide, aud whifh not. y Google V[. SOLID GEOMIlTILV I'KOl'OfcJTIOX SXIV. TUEOKEM 582. Thr sum of nnij fro f>w<' mujU, of a trlhedru! an'jtc is greater than the third faca atKjh. Given the trihedra! angle. S-ABC, with angle A8C its greatest face angle. To prove lASB + I HSC gvaai-n' {linn lASC. Proof. Ill the faee AS(^ ihviw n;>, itiakiiig ZASJ)^ ZASli. Take *7>=,S7>'. In the faee ASC draw the Hne AUC in any convenient direction, and draw AB and BV. Then, in the A A8B and ASB, SA^SA. SB = SD, and Z,l«fi= ZAS'I). :. A A8B=A ASVf. .-. AB=AD. Also AB + BC> AC. Hence, subtracting the equals AB and AJ). BC> BC. Hence, in the A BSC and DSC, 80=^ 8C, «B=Si>, and BC > DC. (Why!) .-. /. BSC is greater than Z DSC. Art. 108. To each of these uneqnals add the equals ZASB and ZASD. :. ZASB+ :/ B«0 is greater than I ASC. (Whyt) 0. E- D. Ex. Iq the above figure, if IAj'<C equals one of the oilier £ace angles at S, as lASB, bow is the theorem proYod J (Why!) (Wl,y=) (my ?) (Why?) (\Vby?) (Why!) y Google POLYHEDKAL ANGLES o5d Proposition XSV. Theorem 588. The sum of the face angles of any convex polyhedral angle is less than four right angles. Given i.l!c polyhedral angle S-ABCDE. To prove the sura of the face ^ at 5 less than 4 rt. ^ . Proof. Pass a plane cutting the edges of the given poly- hedral angle in the points A, B, C, J), E. Prom any point in the polygon ABODE draw OA. OB, OC, oh, OE. Denote the A having the common vertex S as the S A, atid those having the common vertex O as the A. Then the sum of the A of the S A=^ the sum of d of the A. AH, 11)4. Bat ZSR'l+/S7J0iagi-eatert!ian lABC, 1 , . „„ Z5C/J+ ZSCiMsgreaterthan ^BOT, etc.) ' •'. the sum of the base <i of the S A > the snra of the tase A of the A. Ajt, 9. .■. the sum of the vertex d of the S A < the sum of Uie vertex A of the A, As. ii. {if unequal s be suhtratilid from ctiiiiih, Ihe remainders are iiiicqiinl iiirermeor'Ui). But the sum of the d at 0=4 li. i . (Wiiy?) .'. the sum of face d at ii < 4 I't. A . Ax. s. y Google <i04 BOOK V!. SOLID GEOMKTRY PnOPOSITlOX XXVl. TlUXlREM 584. JS iii-0 Irihedml ainjhs luu-c ihe thnc face migh^, of one eqmtl to the tliree fuce uiigl'-s of ilie other, ike (yi- hedral angles have their correspiinding dilwdml niifjUsequn}. and are- edlher equal or sijmmptriruJ . iircordiiir/ as fl/d,- corresponding face angles are arraniU'd in, f/w mine or Ui reverse order. Given thf trihedral i ^-ABC and ^'-A'B'C, liavin- the face A AUB, ASO 3.\\<\ BWC'equal to the face A A'ii'li'. A'S'C and B'S'C, respectively. To prove that the con-espon<3ing dihedral A of S~ABf' and S'-A'B'C are equal, and that A S-ABO and N'- A'B'C are either equal or symmetrical. Proof. On the edges of the trihedral A take ^S'.l, .S7>', 8C, S'A', 8'B', S'C ail equal. Draw AB, AC, EC, A'B', A'C, B'C. Then, 1, In the A ASB and A'S'B', SA = S'A', SB = S'B', and AASB= ZA'S'B'. (Why?) .-. A ASB^A A'S'B'. (Why!) .-. AB^A'B'. (Why?) 2. la like manner AO=A'C', and HC = B'C". :. A ABC^A A'B'C. fWhyf) yGoosle EXERCISES OS THE LINE AND PLAKE 355 3. Take D a convenient point in SA, and draw DE in the face ASB, and DF in the face ASC, each ± SA. DE and i>F meet AB and .10 in points E and JP, respeetivelv. Similarly, take S'lf^SD and construct A J>'l-yr'. Then, in the rt. ^ ADE and A'B'E', Alt^A'l)', and Zi>jli7-./rA'l,". (Why?) .-. A ;inj;-A A'D'E'. (Why?) .-. A/^'=^'i", !ind !)E=I)'E'. (WliyM 4. In like manner it m:\y\K sho^yu that AF=A'F', and .-. A .4Z:F=A J.'£'F'. (Wii;-?) And EF^E'F' (Whyf) 5. Hence, in the A Z)i;F and B'E'F', T>E=n'E', I>F = D'F and EF=E'F'. (Why ?) .-. A DEF^A D'E'F'. (Why f) .-. IEJ}F=IF:J)'F. (Why?) Bat these i are the plane i of the dilieitral A whoso edges are S.l and N'A'. .-.dihedral Zif-Afr-C^diliedrnl IB'-A'S'-C Art. 5.12. In like manner it may be shown that the dihedral A at SB and S'B' are equal; and that those at KO and S'C are equal, .'. the trihedral A S and S' are either equal or sym- metrical. ^^'^^- 5'9' ''^o. Q. E. B. EXERCISES. CROUP 6't THEOREMS CONCEKNIXG TUE LINE A\D f'LANE IN SPACE Ex. 1. A segment of a liiie not parallel to a plane is longer thira its projection in the plane, Ex. 2. Equal stE-aight lines drawn from a point to a place are equally incliiit-a to the plane. y Google 35G B Ex. 3. A line parallel. Ex. 4. If thre perpendicular to a SOt-in GEOMETRY H perpeudiculiir to the si ig in three stniight lines acs £ iulerscction Jii't parallel. Ejt. B. If a plane bisects any line at riglit angles, any point in the plane is equidistant from the ends of the line. Ex. 6. Given JIl X plane .If-V, and ,)(.' ± piano SS; prove JIC ± yii. Ex. 7. Given PQ X plane .1/-V, PR X plane HL, and BS 1 plane J/A'; prove QS X AB. Ex. 8. If a line is perpendicular to one of two intersecting planes,. its pro.jpctioQ on the other plane is perpeniJiQuliir to the line of intersection of the two planes. Ex. 9. Given CE X BE, AE I T>E, and Z C-AD-E a rt. dihedral Z prove CA X plane DAE. Ex. 10. The projections of two parallp a plane are parallel. (Is the converse theorem also true ?) Ex. 11. If two parallel planes are cut hy two non -parallel planes, the two lines of intersection in each of the parallel planes will make equal angles. Ex. 12. It a line is perpendicular to a plane, any plane parallel to the line is perpendicular to the plane. (Is the converse true !) Ex. 13. In the figi DC; prove BF X DC. Prop. VI, given AFi X MN and AF X X DC. Ex. 14. Two planes parallel to a third plane are parallel to each [Sua. Draw a line X third piano.] y Google EXERCISES ON THE LIKE AND PLASB Ei. 15. straight liii The projoetioiis up 9 equal nnd p^r on a plane of t illel. Ex. 16. section. Al ne parallel to t vo pluues is p Ki. 17. prove the a la ngle the flguro to Py.p. XXII. it JRi' obtuse. Ei. 18. In a quadrilateral in apace (i. e., a quadrilateral whose vertices are not all in the same plane), show that the lines .ioliiing thu midpoints of the sides form u parallelogram. Ex. 19, the oppoait The esid lines joiniug the midpoints o s of a quadrilateral in Space Ij: Ex. 20. three faces The planes biseeti Hue every po g the ilihedral ut of wliii;h is [SUQ, Sbo. rt. m2.] Ex. 21 7 n 0(J bisecting I ROS, PQ X plane EOS, QR X OR, QS X OS; vo PIl=PS, PR X OR, □ii PS X OS. ■qual and par pavallel to their ii Ex, 22. In n plane bisecting a piveu plane nn^^e, and perpendicu- lar to its plane, every point ia equidistant from the sides of the angle. [Sl'q. See Ei. 21 ; or through P any point in the bisecting plane pass pianes X to the aides of the iL , etc.] Ex. 23. In a trihedral angle, the three planea bisecting the three face angles at right angles to their respective planea, interaeet iu a line every point of which is equidistant from the three edges of the tri- hedral angle. Ex. 24. If two faceangleaof a trihedral augle are equal, the dihe- dral angles opposite them are equal. Ex, 25, In the figure to Prop, XXIV, provi) that lASC-'r IBSC ia greater than Z ASl) -{- Z ISSU. Ex. 26. The common perpendicular to two lines in spui^e is tho shortest line between thero. y Google '■^A vwr\ i;noi; vt, rood . Given .l/.Vll Jin, and A AIIC => A iibc. trihedral angles are equal. El. 29, Any two sjmmelrieiii trih, Aval angles are equivalent. [Suo. Take S.l, Sli, sr, S'a'. S'li', S'C', all fqiml. Pass planes -IfiC, A'B'C. Draw SO aiid S'ty ± tlieso plitnes. Tben thw trihedral A am divided into three pairs of isoseelos symtuetripal trihedral d , etc.] EXERCISES. CROUP 68 LOCI IN SPACE Find the loeiis ol a point equidistant from Ex. 1. Two parallel planes. Ex. 3. Three t;ivea points. Ex. 2. Two given points. Ex. 4. Two interseoting linei Ex, 5. The three faces of a trihedral angle, Ex. 6. The three edges of a trihedral angle. Find the ioeus Ex. 7. Of all linos passing tlirongh a si^en point and parallel L plven plaice. Ex. 8, Of all liuPB pcrptiKiicul ar to a given line It a given -point n the line. Ex. 9. Of all points in a give n plani 3 eqnidisti mt from a given loint outside the plane. Ex. 10. Of all points equidista ut froii 1 two £ivei 1 points and from ,wo parallel pianos. Ex. 11. Of all points equidista lit frou 1 two give] 11 points and from Swo intersecting planes. Es, 12. Of all poinis at a give ■n dista iicu from a given plane aod y Google EXERCISES OK THE LINE AND PLANE 6Di} EXERCISES. CROUP 66 riiOllLKMS CONCERNING THE POINT, LINE AND PLANE IN SPACE Ex. 1. Through a given point pass a plane parallel to a given plane. Ex. 2, Through a given point [lass a plane perpsniiii:iilai' to a given plane. Ex, 3. Tlirougii a given point to (construct a plani' paralli'l to two given lineR which are nol in the suiiie plai^f. I'rovf that only oui! piaue oun be coiistrurli^il fiilfimii!; Iho givi^n conditions, Ex. 4, Bisect a given iliheilral angle. Ex. 5. Draw a plane equallj inclined to three lines whleh meet at a point. Ex, 6. Through a piven point draw a line parallel to two givan intersecting planes. Ex. 7. Find a point in a plane .such that lines drawn to it from two given points without the plane make equal angles with the plane, [SuQ, See Ex. 23, p. 176.] Ex, 8, Find a point in a given line equidistant from two given points. Ex, 9. Fhui a point in a plane equidistant from three given points. Ex. 10, Find a point equidistant from four given points not in a plane, Ex. 11. Through a given point draw a line which shall intersect [Sua. Pass a, plana through the given point and one of the given lines, and pass another plane through the given point and the other given line, etc.] Ex. 12, Through a given point pass a piano cutting the edges of a tetrahedral angle so that tlie section shall be a parallelogram. fSVG. Produce each pair of opposite faces to interseet in a Straight line, etc.] y Google Book VII rOLYL{EDRONS 585. A polyhedron is i ^lolid buiimlnl \i\ pUnes 586. Tha faces o£ a pohlie 6vou ar(i itri houmliiiy: pfiiiiLS tin, edges of a polyliedrun are tlii. hne% i of intersect ion of its faces A diagonal of a polyhedion lo i a straight line joining two ot ita vertices whicli are not m the j same face. The vertices of i polv- hedrou are the points in wliiji iti, edges meet or intersect. 587. A convex polyhedron is a i>olylK'dvo!i in which a, section made hy any phme is a convex polygon. Only convex polyhedrons are to be considered in Ibis book. 588. Classification of polyhedrons. Polyliedrons are simetimi'i elibsififd tttoidm^ to th mimlei of their fices Thus a tetrahedron it, a pohhedion ot foui fices T hexahedron is a pohh^dion of -ii^ fn. ^ in octahedron 1 n rf tiglt I dodecahedron one if tftcl\c ml an cosabedron one ot twcntj f^ces y Google POLYHEDKOXS The polyhedronB moat important in practiea! life are those deter- mined by their stability, the facility with which thijy can be made out of eommoQ raateviuls, a3 wood and iron, tha readiiieaa with which they can be packed togethti', etc, Thus, pfism meaua "aomething Bawed oil." PEISMS AHD PARALLELOPIPBDS 589. A prism is a polyhedron bounded bj two parullcl planes acd a group of planes whose Hues of iiitersectiou ui "" parallel. 590. Tlie bases of a prism are tbe faces formed by tbe two parallel planet., tbe lateral faces are the faces formed h-\ the group of planes whose lines of inteiseetKm aie paiallel. The altitude of a prism is the perpendicular distance between the planes of its bases. The lateral area of a pi-ism is the sum of the areas of the lateral faces. 591. Properties of a prism inferred immediately. 1. The. lateral edges of a prism are equal, for they are parallel lines iucluded between parallel planes (Art. 589) and are therefore equal (Ai-t. 532). 2. The lateral faces of a prism are parallelograms (Art. 160), for their sides formed by the lateral edges are equal and parallel. 3. The bases of a prism are equal polygons, for their homologous sides are equal and parallel, each to each, (being opposite sides of a parallelogram), and their liomoi- o?ous angles are equal (Art. 538). 592. A right section of a plane perpendicular to thu hit' priM section made by a y Google 3li2 1;00K VII. SOLID ceometisy C.O3 \ trianiFiilnr nri«m \< n iirUin ivlio';.' liilKf n I it 594. All oblique prism is ii prism wliose Literal edges i-e ol)lir|iie to the liases. 595. A right prism is a prism whose lateral {idgcs are pei'pemlieiitar to the bases. A regular prism is a right prism f ses are regiilai- polygons. 1 whose bases 597. A truncated prism is that part <if a pvisiii included between a base and a section nnide by a plane obliiine tu llie base and cutting all tlie lateral edges. 598 \ parallelo piped j wl e ii II n r I'll ' " ' 599. A right parallelepiped is a purallelopiped whose lateral edges are perpendicular to tiie bases, y Google PmSJtS AND rAKALLELOPIPEDS 363 600. A rectangular parallelopiped is a right pavuUelo- piped whose bases are rectanglos. Hence, all the fitces of a rcciuiujidur puralldojyipcd are rectanglns . 601. A cube is a rectangulai- parallelopiped whose edges are all equal. Hence, all the faces of a ciihe are s(iHares. 602. Tlio unit of volume is a eube whose edffe is equal to some linear unit, as a cubic inch, a, cubic foot, etc. 603. The volume of a solid is the uuraber of units of volume which the solid contains. Being a iiwHiber, a volume may often be dptermitied from other numbers in certain expeditious wiiys, wliioii it is one of tlie objects of geometry to determine. 604. Equivalent solids are solids whose volumes are equal. it number of faces whiol El. 2. A square right prism is what JEind of a parallelopiped f Ex. 8. Are there more ri^ht parallelepipeds or rectangular paz'al- lelopipedst That is, which of tliese indiides tUe other as a special <;aser Ex. 4. Prove that it a given straight line is perpeinlicular to a Ten plane, and auotber straiffht linn is perpendicular to another i-aae, aad the two planes are parallel, Iheu the two given lines ara y Google BOOK VII. SOLID GEOilETHY Proposition I. Tiieoiiem 605. Serliona of a piiam vtadc hij pantUd planes cuUing all the lateral edges are equal pohjijvns. Given tlie prism PQ out by || planes forming the sections AB and A'B'. To prove section ^7J = secttion A'B'.- Proof. AB, BC, CI), etc., arc || A'B', B'C, G'ly, etc., respectively. Art. 53i. .-. AB, BC, CI), etc., are equal to A'B', B'C, C'D'^ etc., respectively. Art.isr. Also A ABC, BCD, etc., nre eqiuil to AA'B'C, B'L'l)', etc., respectively. Art. 538. .-. ABCDE^A'B'C'D'E', Art. 47. Q. E. B. 606. Cor. 1. livery section of a prism made by a plane parallel to Ike base is equal to the base. 607. Con. 2. All fiylU sneliuns of a pt-ism are equal. y Google PRISMS 365 Proposition II. Theorem 608. The lateral area of a prism is equal to the product of the perimeter of a right section hy a lateral edge. BX^aCJ GJ^BCXE, area £17 IQ = CT>XE, etc. Given the prism KQ, with its lateral area iloiioteil by 8 and lateral edge by E; and AD a right section of the givou prism with its perimeter denoted by P. To prove S^PXE. Proof. In the prism liQ, each lateral edge = /7, AiLTjOI, i. Also AB ± GH, BG X IJ, etc. Art. 505. Hence a.vez.CJ RE^AB X GE^AB X ^,1 ■ni^' ^ V Art. ?.Sr>. But 8, the lateral area of the prism, equals thi> sum of the areas of the ZX7 forming the lateral surface. .-. adding, 8'^UB + BG+ CV + etc.) X E. Ax. 2. Or S^PXE. q. E. D. 609. Cor, The lateral area of a right prism eQv,als the product of the perimeter of the b<tse hy the altitude. Ex, Find the lateral area of a rit'lit prtam whose nltitiide U 12 in., and whose buso is an equilateral triangle witti a side of G in, Alsa flad the total area of this lijfuro. y Google otlfl BOOK Vn. SOLID GEO^IETItY PiiorosiTiox III. Theorem 610. If in-o pnf.ms Jiave the lliirp fa-'cs mcludln// a trihi'iJrid angle of one eqmd, rf^pectireli/, to the three faces iiidiidhifi a trihedral angle of the oOier, and similarly placed, the prisms are equal. Given tlie prisms AJ and A'J', having the faces AS, AJ>, AG equal to the faces A'K', A'D', A'G', respectively, suA similai'ly placed. To prove AJ=A'J'. Proof. The face d EAF, EAB and BAF arts equal, respectively, to the face d E'A'F', E'A'B' and B'A'I". Hyp. .-. trihedral ZA = trihedral AA' Art. 584. Apply the prism A'J" to the prism AJ, makine each of the faces of the trihedral /.A' coincide with corresponding equal face of the trihedral LA. Geom.Ai.2. .". the plane F'J' ivill coincide in position witJi the plane FJ. Art. 500. {/"I- tliepnbils C, P, K' coincide Kith G, F, K, respectively). Also the point C will coincide with the point C. ,', Cfl'wili take the direction of CH. Geom.Ax. 3. .'. S' will coincide with H. Art. 508, i. In like manner J' will coincide with J. Hence the prisms AJ and A'J' coincide in all points. .-. AJ=A'J'. Art, 4T. 611. Cor. 1. Two truncated prisms are equal if the ihrep. faces i)icliiding a trihedral angle of one are equal to the thr&e faces including a trihedral angle of the other. y Google 612. COK. 2. Two right prmns are equal if they have egiMii bases and equal altitudes. Proposition IV. Theorem 613. An oblique priavi is equivalent to a riqht prism whose base is a right section of the oblique prism and whose altitude is equal to a lateral edge of the oblique prism. Given the 'oblique pvism AD', with the right section JV"; also the rigrht priKin FJ" whose lateral edges are each equal to a latei-al edge of Alf. To prove AD' ^FJ'. Proof. AA' = FF. Hyp. Subtracting FA' from each of these, AF=A'F'. ("Why ?) Similarly BG = B'(}'. Also An = A'B', and FO^FO'. Art. 155. And A of fare iH7 = homologous A of face A'G'. Art, 130. .-. face AW^f.iee A'G', Art 47. (fur ihey haco all tlieir parts eijuul, each to each, and .'. can he made In like manner face -iK^=face A'K'. Eat faee Ai) = Eace A'ly. Art. 5B1, 3. .■. truncated priam j4J=tninc.ated i)rLsm A'J'. Art, oil. To each of these equals add the solid i-7/. .-. AB'^FJ'. (Why?) Q.B. ». y Google 3G8 EOOK VII. SOLID GEOJIETUY PfiOTOSITlON V, ThEOEEJI 614. The oppo^Ht Uiferal facfs of a paralli'Iop-ipcd are equal U)id parallel. Given the piiralleSo piped AH with tiie base AG. To prove AG = an.d \] EH, anil AJ= aud || BH. Proof. The base AC is a /ZJ . Art. 598. .-. An = am\ !! EC. (Wiiy?) Also the lateral face AJ is a CH . Art. 591, 2. .-. ^F^and || EJ. (Whjf) .-. inAF=ZCEJ. Art. 539. Aua £Z7AG^/-y EH. Art. m. Also pUne A(7 11 plane EH. Art. 538. In like manner it inay he shoivii thitt AJ and HH are eqnal and parallel. 0. E. D. 615. Con. All// tiro oppoi^ite faces of a pandlelopiped may be lalrn «n the ba.'ics. Ex. I. How roBny edKes has a paruHelopiped ! How many faoea ? How many dihedral angles ? How many trihedral angles f Ex. 2. Find the lateral area of a prism whose lateral edge is 10 Ex. 3. Find the lateral area of a right prisni whose lateral edge >a IG and whose base is a rhombus with diagonals of 6 and 8 in. y Google Proposition VI. Theorem 616. A plrnie passed throngli liro didgiyii'dhj opposite edges of a paralklopiped divides the parallelopiped into two equivalent triangular prisms. Given the parallelopiped AE with a plane passed through the diagooally opposite edges AF and CS, forming the tri- angular prisms ABO-G and ABO-E. To prove ABC-G^^ADC-K. Proof. Construct a plane X to one of the edges of the prism forming the right section PQRS, having the diagonal PH formed hy the intersection of the plane FHCA. Then PQ 11 Sn, and QR \\ PH. (WhyT) .-. PQRS is a ZZ7 . (Why ?} .-. A PQB= A PSE. (Why ?) But (lie triangular prism J.BC-0 ~ a prism wlinse hase is the right section PQR and wliose altitude is AP. Art.6i3. Also the triangular prism ADG-K o a prism whose base 's the riglit section PSIi and whose altitude is AF. (Whyt) But the prisms having the eqnal bases, PQE and P8R, SQd the same altitude, AF, are eqnal. Art, G13. .-. ABC'G ^ AD(J-K. A3 1. q. E. ». y Google dii) BOOK VII. SOLID Gr.o:^n:TiiY Proposition VII, ThEOI!EJI 617. If two rectangular parallelopipitiWImvii equal bases thru "'■^ ^0 each other as their allihules. Given the rcctaiiguhir parallnlo|Hpi.';lB P' and P having equal bases and tlie dtitutles A' li' ami AH. To prove P' -. P^A'B' -. AD. Case I. When ike altitudes AT/ iniil AP, 'are com- mensurable. Proof. Find 'a eoimnon measure of A'li' and AB, aa AK, and let it lie contained in A'Ji' n tiinos and in A}i m times. Then A'B'_: AB^n -. vi. Tlii'ough the points of division of A'B' and AP> pa>,3 planes parallel to the bases. These planes will divide I" into ii, and P into m small reetansjukr parallelopipeds, all erpml. Ai-t. 612. :. r : P = n: m. .: F : P==A'B' : AB. {Why ^ Case II. When the altitudes A'B' and AB are incoin- mcn.'iurable . Lut the pupil supply the proof, using the method of limits. (See Art. 554). 618, Def. The dimensions of a rectangular parallele- piped are the three edges whieh meet at one vertex. y Google l'AKAI.LELOFIPEDS 619. Cor. If two rectangular partiUelopipeds haw two dimensions hi common, they are to each other as their third dimensions. PHOPOSITION VIII. TlIEORRM 620. Two rectangular ■jxirallclo'pipcds having equal alti- tudes are to each other as their bases. Given the rectangular parallelopipeds P and P' having the common altitude a, and the dimensions of their hases 6, c and &', c', respectively. P ^ hXe P' b'X c' Proof. C«iii^tri.R't 1 he rectangular parallulopiped, <i>, whose altitude isit and the dimensions of who^e base are h and c'. To prove Then Moltiplj'ing the corresponding mciubers of thes equalities, 621. Cob. Two rectangulay paralldopipedH having one 'U'ltiension in common un- to each other as the products of the other iiro diiiicnsioiis. y Google iSiJ BOOK Vn. SOLH) liEOMIiTKY Proposition IX. Theorem 622. Ai"j Ih-o rpciangulnr pfirnllelopipeds are to each other as tlw pro'liic/a of ilinr thne dhi Given tht! rectaiigulai- parailuloplpeds Paiid 2" havir the dimensions «, h, c and a', V, e', respectively. To prove ' X fi' X >:' Proof. Coiistnict the reetaugular panillolo piped Q hav- ing the dimensions a, li, c'. P r. Then 7}^~' Art. cis. Multiplying the corresponding members of these equalities, P_ » X&X c P a'Xb'Xc'' ^''*' Q- E. ». Ex. 1 . Find the ratio of the volumes of two reetaagular paraOelo- pipeds whose edges are 5, G, 7 in. and 7, 8, 9 in, Ex. 2. Which will hold mote, a bin 10 s li 1 7 ft., or one Sx 4 s 5 ft. 1 Ex. 3. How miiuy Liicks 8x4x3 in. are neeesaary to build a wall 80s6 ft. sH ja. ! y Google PAliALLELOPirEDS Proposition X. Theorem 623. The volume of a rectangular paralleJopiped is eqtial the product of itn three dimensions. Given the rectangular para lie loin [jed F having the three dimensions a, 6, c. To prove volume of P~a X 6 X c. Proof, Take as the unit of volame the cube C, whose edge is a linear unit. Then - = aXbXc ^ U 1X1X1 The volnoie of P is the number of times P containK the unit of volume U, or — Art. 603. .". volume of P=« X 6 X i;. 624. Cor. 1. The volume of a cube is the ciihe of its edge. 625. Cor. 2. TM vobime of a rectangular paraUelo- piped is equal to the product of ita base by its altitude. Ex, 1. Find the numljer of cubie iuchtg in the volume of a cube whose edge is 1 ft. 3 iu. How many bushels dots this box contain, it 1 bttshel=2150,42ou. in. t Ex. 2. Tlie meaBurempiit ot tiie volume if a OAibe reduces to the it of the length of what aiiigiu striiiglit line f y Google 374 B0(1K VII . SOLIll GEOMETRY PROPOSITION XI. THBOai:.1I 626. The volume of any pamllelopiped is' equal to th^ product of its base by its altitude. Given the oblique pai-al!t?lopiped P, with its base denoted by B, and its altitude by H. To prove volmne of P=7J X E. Proof, In P produce the edge CD and all the edges parallel to CD. Ou CJ) produced take FQ^CB. Pass planes through F and G ± the produced edges, forming the parallelepiped Q, wltli the rectangular base denoted by B'. Similariy produce the edge GI and all the edges !1 GL. Take IK=OI, and pass planes through I and A' X the edges last prodneed, forming the rectangular parallclopiped B, with its base denoted by B". Tlion P. >e^ >R. Art, G13 Also £< sP' -P". Art. 3S6 But volume ofB .B XP. Art. 625 .'. volume ofP- .B' Xff. Ax. 1 Or volume [OutliEe Proof. F OtP =.« = -P X U. S^B" KM. = P Xil.] Ax. 8 «. B. P. y Google Proposition XII. Theorem 627. The volume of a triangnlur prism is equal to the product of Us hu^e by its aUitude, Given the triangalar prism PQB-M, witli its volume denoted by 1', area of base by B, and altitude by JI. To prove Y=BX II. PQ, Qii. Q^J^, construct the Proof. ITpou the edg parallelopiped QK. Hence (^ff^twice FQH-M. But volume of QK=in-iin PQBTX R. = 2B X H. .: twice volume PQR-M==2B X H. :. volume PQRSI^ B X H. Ex. 2 Find the volumo o! a and tLo edata ol whose base ar y Google 376 HOOK VII. SOT.TD GK(DMETHY PROPOSITION XIII. Theorem 628. TJie fnltinie of (inij prism is vqual to the produU of Us base bi/ its altitude. Given tlie prism AK, with its vohime dmioted by V, area of base by B, and altitude by H. To prove V=BXE. Proof. Through any lateral edge, as AF, and the dlag- onaJs of the base, AG and ylJ), drawn from its foot, pass planes. These planes will divide the prism iuto triangular prisms. Then V, the volume of the prism ,-17i, equals the sum of the volumes of the triaugnlar prisms. As. G. But the volume of each triangular prisni = ita base X 11. Hence the sum of the vohiiues of the A prisms = the sum of the bases of the A prisms X II. = liy.U. Ax. 8. .-. V=BXE. kn. 1. <}. E, D, 629. Cor. 1. Two prisms are to each other as the pro- ducts of their bases by their altitudes ; prisms having equiva- lent bases and equal altitudes are equivalent. 630. Cob. 2. Prisms having equivalent bases are to each other as their altitudes; prisms having equal altitudes are to each other as their bases. y Google 377 PYRAMIDS 681. A pyramid is a poiyhcdrou bounded by a. group o£ planes passing through a common point, and by another plane cutting all the planes of the group. 632. The base of a pyramid is the face formed by the cutting plane; the ' lateral faces are t!ic faces formed by the group of planes passing through a com- mon point; the vertex in the common point through wliii-h the group of planes passes; the lateral edges are the inter- sections of the lateral faces. The altitude of a pyramid is the perpeodicular from the vertex to the plane of the base. The lateral area is the sum of the areas of the lateral faces. 638. Properties of pyramids inferred immediately. 1. The hdn-al fan:^ of a pi/nuiiid arc Iriau'jU'K {Art. 508, 2) . 2, Tlie haufi of a pyramid is apnlijgoii (Art. 50S, 2). 634. A triangular pyramid is a pyramid whose base is a triangle; a quad angular py am'd 'a '1 lose base is a quadrila A triangular pyram a d b U faces. All these fa a ay miuya taken aa tha basf. 635. A regu ar pyram d is a pyramid who b regular polygon, ud fo of whose altitu with the wiiLcr cf h b y Google 378 BOOK VII. SOIJD GEOJIirSRY 636. properties of a regular pyramid inferred immedi- ately. 1. T]te hiieral ei^yc.'i of a rajidur pymmkl are equal, for they are oblique lines drawn from a point to a plam; cutting off eqnal distances from tlie foot of the per- pendicular from the point to the plane (Art. 518) . 2. The lateral fwes of a regular pi/ranud are cqiiaj igifs- celes Mangles. 637. The slant height of a regular pyramid is the ahl- tude of any one of its lateral faci's. The axis of a regnlar pyramid is-its aititode. 638. A truncated pyramid is thu portion of a pyramid included between the base and a section cutting all the lateral edges. 639. A frustum of a pyramid is the /fMT^ part of a pyramid included between the //M^-lw base and a place parallel to the base. \gl.-...>J"' The altitude of a frustum of a pyramid is the perpendicular distance between the planes of its bases. 640. Properties of a frustum of a pyramid inferred Im- mediately. 1. the lahral faces uf a frasiam of a pyramid are ira peloids. 2. The lateral facet! of a frnglnm of a rerjulur punnnid are equal isosceles trapesoids. 641. The slant height of the frustum of a regular pyra- mid is the altitude of one of its lateral faces. Ex. 1. Sliow that tlie foot of tlie nltitude of a regular pyramid eoineides with the center of the cirele eircumscribed abont the base. Ex. 2. The perimeter of the midaeetlon of the [rustum of a pyra- mid equals oue-iuilf tha aura of the potimutm'a of tlie bases, y Google rvRAMiDS 379 Pboposition XIV. Theorem 642. The lateral area of a regttlar pyramid is equal io half the product of (he slant height hij the perimeter of the Given 0~ABCI)F a regukr pyi'amkl with its lateral area denoted by S, slant heiglit by L, and perimeter of its base by P. To prove g=J£XP. Proof. The lateral faces OAJi, O/iO, etc., are equal isosceles A. Art. GoG, 3. Hence each lateral face has the same slant height, L. :. the area of each lateral face = -} L X its base. .-. the sum of all the lateral faees = i L X sum of bases. ^hLX P. :. *S' = i LX P. A^,8, 643. CoK. 77te Intend urea of thefrustnm of a regular puraviid is equal io one-half the mm of the perivieters of its iases multiplied by its slant height. Ex. Find tlie lttterii.1 ii lieifj'lit is 32, uiid an l-S uvea liXici, I of whosi; base ia lli. Piad tho 1 y Google ]iOOK VII. SO"LtI) OEOJIETltY Pkopositiom XV. Theokem 644. If a pyramid is cut by a j)?D«e paruUi'l to the fi«se, I, The lateral edges and the altitude are divided pro- poriionalbj; II. Tke section is a polygon similar to the base. Given the pyramid S-ABCDF, with the altitude SO cut by a plane MN, which is parallel to the base and intersects the lateral edges in a, b, c, d, f and the altitude in o. So To prove I. — =^ 11. 8 A SB SC The section ABCDF. SO abcdf similar to the base Proof, I. Pass a plane through the vertes S II MN. Then SA, SB, SC . . . SO, are lines intersected by three I planes. Sa _Sb _Sc _ ___ S SA SB SO SO Art. 539. II. {Wiiy f ) ab II AB. .: A Sab aud SAB are similar. In like manner the A Sbc, Scd, etc., are similar to the SBC, SCD, etc., respectively. \sb)^ AB ~BC Ksc) yGoosle oSrf/ ABCDF SA^SO " «6" cihcif . . &>' PYRAMIDS 381 That is, the homologous sides of abcdf and ABCDF ave proportional. Also ^afjc = Z ABC, Ihal = I BCD, etc. Art. 538. .•. section aferf/ is similar to tlio buse ABCDF. Art. 32!. Q. E. D. 645. OoE. 1. A section of a pyramid parallel to the hose is to the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. But Ji:^^^ = ABGDF SO- 646. CoE. 2. If two pyramids having equal altitudes are cut hy a plane parallel to their bases at equal distances from the vertices, the sections have the same ratio as the bases. Let S-ABCDF and V-PQE be two pyramids cut as described. n,, abcdf Ho' , , , par Yi Then ^— =;=r- Art. G45 ; also -i-i— = ^^- WhjT ABCDF «o- DQR VT But rr=SO, and n = So. (Why?) . »&«?/ ^ W ohrdf_ABCDF •■ABCDF PQli °^ pqr PQR ^ " ' 647. Cor. 3. If two 2>yrHmids have equal altitudes and 'equivalent bases, seetions iiiiide hy phnies parallel to the bases at equal distances f rum Ihv verlk-vs art equivuhnt. y Google KOLID (iEOMETRY PhOI'OSITION XVI. TlIKOREM 648 Tlifi lohaiic of k tn mgitlai j^yrmmd is the limit of i Hin "I the loliiwcs of n sciipf of inscribed, or of a series f iKtni''' hIkcI ;;iisws of qtml altitude, if the number of hi. tndif.miibj inueat,€i^ Given the triaogiilar prism 0-ABG with a series of in- scriijed, and also a series of circnmseribed prisms, formed by passing planes which divide the altitude iuto equal paris, aud by making the sections so formed first upper bases, then lower bases, of prisms limited by the next parallel plane. To prove O-ABG the limit of the sum of each series, if the number of prisms in each be indefinitely increased. Proof, Each inseribad prism equals the circumscribed prism immediately above it. Art. 629, .■. {sum of circnmseribed prisms) — (sum of inscribed prisms) = lowest civcumscribed prism, or ABC-K. If the number of prisms be indefinitely increased, the altitiide of each approaches zero as a limit. Hence volume ABG-K=0, Art. 253, 3. {for Us base, ABC, is cojistant Khile Us altitMe = ) . .■, (snm of circumscribed prisms) — (sum of inscribeil prisms) = 0. .■. volume 0~ABC — (either series of prisms) ^ 0. {for this difference < (ajyereiice between the two series, vihick lust difference = 0). ,■, 0-ABC is the limit of the sum of the volumes ot either series of prisms. q. e. b. y Google Proposition XVII. Theorem 649. If two triangular pyramids have equal altitudes and equivalent bases, tJiey arc equivalent. Given the triangular pyramids 0-AEG and 0'~A'B'C' having equivSreQt bases ABC and A'Ji'C, and equal altitudes. To prove O-ABC'o.O'-A'B'C. Proof. Place the pyramids so that they have the com- inou altitude R, and divide H into any convenient nnm- ber of equal parts. Through the points of division and parallel to the plane of the bases of the pyramids, pass planes cutting the pyramids. Using the sections so formed as upper bases, inscribe a series of prisms in each pyramid, and denote the volumes of the two series of prisms by V and I". The sections formed by each plane, as KLM and K'l/M', are equivalent. Art. 6*7. .'. each prism in 0-ABC o corresponding prism in O'-A'B'C (Art, C29). .-. F= F. As:. 2. Let the number of parts into which the altitude is divided be increased indefinitely. Then V and V become variables with (h-ABC and 0'~A'B'C' aa their respective limits. Art. MM. But I'oT'' always. (WLv?) :, 0-ABC~0'-A'B'C'. (Why!) (1. X. D. y Google BOOK V)I. SOLID PHOl'DSITrON XVIII. TlIEORF,:« 650. Thr vohinK' of a fridDditlar pyramid i oiif-lhird Ihr iirodiid of iV.v bam: hy il^ nUilmh'. Given the triangiilar pyramid 0-AB€, having its volnmo denoted bj- V, the area of its base by B, and its altitude by H. To prove V^^BXM. Proof. Ou ABC as a base, with OB as a lateral edge, construct the prism ABG-BOF. Then this prism will be composed of the original pjTa- mid 0-ABG and the quadrangular pyramid 0-ADFO. Through the edges OJ) and OC pass a plane intersecting the face ^DFC in the line B<7, and dividing the quadrangular pyramid into the triangnlar pyramids 0~ADC and OSFC. Then 0-AT)C^O-I>FC. Art. 649. {for IkeijMrc the common rerlci 0,<i,nd the equal bases ABC andBFC). But 0-DFC may be regarded as having as its vertex and DOF as its base. Art. 634. .-. (}'DFC=:^0-ABC. Art. 619. .'. the prism is made up of three equivalent pyramids. .-. (}-ABC=h the prism. As. 5. But volume of prism = B X H. Art. 627, .-. 0-ABC, or r=iBX H. Ax. 5. g. E. B. Ek. FJ^id the volume of a trmngula,r pyramid whosa altitude 13 12 ft,, ami whose b:iao is an equilateral triangle with a bide of 15 ft. y Google Proposition XIX. Theorem 851. The volume of any pyramid is equal to one-third iAe product of itn base by Us ultitude. Given the pyraraid 0- ■il>( VF ln\ mg its volumfi denoted ^>y V, the area of its baae bj £., and its altitude by H. To prove F= i B X E. Proof. Through any lateral edge, as OD, and the diago- nals of the base drawn from its foot, as AS and BD, pass planes dividing the pyramid into triangular pyramids. Then Y, the volume of the pyramid 0-ABCDF, will Bqnal the sura of the volumes of the triangular pyramids. But the volume of each A pyramid = J its base X E. Art. 650. Hence the sum of the volumes of A pyramids=i sum of their bases X E. Ax. 2. = k I) X E. Ax. 8. .-. r=h BX S. Ax. 1. Q. E. D. 852. OoR. 1. 5'fie volumes of two pyramids are to each '>iher as the products of tkeir banes atid altitudes; pyramids fiaving equivalent bases and equal altitudes are equivalent. 658. Cob. 2. Pyramids having equivalent bases are to ^eh other as their altitudes; pyramids having equal alti- 'Mifes are to each other as their bases. 854. Scholium, The volume of any polyhedron maybe J'^und by dioidiuff the polyhedrmi Into pyramid.'^, finding the ^hnie of each pyramid •jcpundchj, and tailing their sum. y Google BOOK VII. aOI.ID GEOMETRY Proposition' XX, Theorem 656. The frustum of a triangular pyramid is eguivaleKt to the sum of three pyramids whose common altitude is the altitude of IM frustum, and tvJioae bases are the lower base the upper base, and a mean proportional between the two bases of the frustum. Given ABG-DEF the frustum of a triangular pyramid, having the area of its lower base denoted by B, the area of its upper base by b, and its altitude by B. To prove ABG-DEF ^ three pyramids whose bases are B, h and V Bb, and whose common altitude is H. Proof. Through E and AG, E and DC, pass planes divid- ing the irustum into three triangular pyramids. Then 1. E-ABC has the base B and the altitude H. 2. E-DFG, that is, G-DEF, has the base b and the altitude if. 3. It remains to show that E-ADC is equivalent to a pyramid having an altitude H, and a base that is a mean proportional between B and 6. Denoting the three pyramids by I, II, III, I_ A ABE __AB^_AC^ ^ ADC ^11 II A AI)t:~DE BE A I>FG III' (Arts, 653, 391, 644, 321. Let the pupil supply the reaaon for each I II step in detail). (Ax. l.)orII = l/lXiny .-. E~ADG=y'{^HXB) (hSX b}=iHVBXb. Hence, J.£OZ'iJ#— sum of three pyramids, as deseribecU ^. £■ S. y Google 656. Formula for volume of frustum of a triangular pyramid. F=i JI {B+h + VmY. Proposition XST. Theorem 657. The volume of Oie fi-usium of nny pyramid is eguivaJent to the sum of the volumes of three pyramids, whose common attitude 4s the altitude of the frifsium, and whose lases are the lower base, the upper base, and a mean propor' tional between the two bases of the frustum. Given the frustum of a pj-ramid Ad, having its volume denoted by V, the area of its lower liase by B, of Its upper ^>ase by 6, aud its altitude by H. To prove T=h E (B + b + VBb). . Proof. Produce the lateral faces of Ad to meet in K. Also construct a triangular pyraiuid with base PQR ■qnivalent to ABCDF, and in the same plane with it, and ■Tith an altitude equal to the altitude of K-rABCDF. Pro- luce the plan€ of. ad to cut. the second pyramid in pqr. Then pqr'cahcdf. Art. 647^ .-. pyramid ST-ABCPFo pyramid T-PQIi. Art. (552. Also pyramid K-abtdf = pyramid T-fqr. (Why ?) Sttbtractiug, frustum Ad ^ frustum Tr. Ax. 3. But volume Pr^-]. 11 Hi Jrh+VlThh :. volume Ad^;, II {n ]-b^VBb).' (Why?) -"Q. EiB. - y Google SOLID GEOMETRY Piioposrnox XXIL Tiieorkm 658. A truncuted triangular priim is vqvivalent to tTie sitvi of three pumHiids, of which the base of the prism is the common base, and whose vertices are the three vertices of the inclined section. Given the truncated triangiilai- prism ABG- PQR. To prove ABC-PQR ^ the sum of the three pyramids P-ABC, Q~ABC and Ti-ABC. Proof. Pass planes through Q and AC,Q and PC, divid- ing the given figure into the three pyramids Q-ABC, Q-APG and Q-FBG. 1. Q-ABG has the required l*ase and the required vertex Q. 2. Q~APC^B-APG, Art.652. (/or they liave tkesame base, AFC, and the same altitude, their ver- iices beind in a line |i bate APC), But B~APG may be regarded as having P for its ver- tes, and ABG fov its base, as desired. Art. 634. 3. Q-PBC<i= B-ARC (see Fig. 2). Art. 652. \foT the base ASC ^ base PSC (Art. 300); and the altitudes of the iao pyramids are eqnal, the vertices Q and B being in line || plane FACB, in lahich the bases lie). But B~ARG may be regarded as having R for its ver- tex, and ABC for its base, as desired. Art. 63* y Google -•. ABC-PQB o sura of three pyramids whose eommoii base is ABG, and whose vertieea are P, Q, R. 659. Cor. 1. The vobinie of a truncated right triangii- iar prism (Fig. -i) '.v '■q/'id to the product of its la^ie hij lyne-third the auin of Ur lateral eOges. 660. Cor. 2. The vohime of any truncated triangular prism (Fig. i) is equal to the product of the area of its ngM section (ii/ one-third the sum of its lateral edges. P8ISMAT0IDS 661, A prismatold is a polyhedron bounded by two polygons in parallel planes, called bases, and by lateral faces which are either triangles, trapezoids or 662. A pfismoid is a prismatoid in *bich the Itiises have the same number of sides and liavt- their corresponding sides parallel . &"■ The Tolurae of a truncated right parallelopiptd equals the •fea ol the lower base multiplied by oue-fourth the sum of tlie lateral ^i^eea (or by a perpend'ciilar from the ceuter of the upper base to tUe lower basul. y Google r,i:inn-ri'i{Y Proposition XXIII. TiiKOiiEJi The vohmie of a prisituifold is iq'int (o the product of its ultiiude hy tin: si fowr Hmes the area of its midsection. sixth of its bases and uf Given the prisiuiitoitl ABCI'-FOK, with bases B and h. midsection Jf, volume 1', atul altitude Tl. To prove Y^\H {B-^h^A 21) . Proof. Take any point in tlie midsection, and through it and each edge of the prismatoid let planes be passed. These planes will divide the figure into parts as follows : 1. A pyramid with vertex 0, base ABCD and altitude i H, and whose volnme .". = ,V 77 X jR. Art. 65] , 2. A pyramid with vertex 0, base FGK, and altitude i if, and whose volume .■. =i 7/ X h. (Why?) 3. Tetrahedrons like O-A'lKl who^e volume may be determined as follows (see Fig. 2): AB^-l-FQ. (Why!) .-.A .it?B=4 A PGQ. Art. 398. .-. O-AGB^i 0-PGQ. Art. 653. But 0~PGQ (or G-FQO) = h I'QO X h H = i HXPQO. .: O-AGB^i SXiAPQO. (Why?) .-. the sum of all tetrahedrons like 0-A(3B=i HXiM. :. Y=-\ EXB + i HXh+i 11X4 M. Or V=}E iB+h + iM). l!.S.l>. yGoosle RFGULAE POLYHEDRONS REGULAR POLYHEDRONS 664. Def. a regular polyhedron is a polyhedron all of whose faces are equal regular polygons, and all of whose polyhedral angles are equal. Thus, the uuhe is a regular polyhedron . Proposition XXIV. Theorem 665. Btit five regiihir polyh^dronti are p Given regular polygons of 3, 4, 5, etc., s To prove that regular polygons of the same number of sides ean be joined to form polyhedral <^ of a regular polyhedron in but five different ways, and that, conse- quently, but five regular polyhedrons are possible. Proof. The sum of the face A of any polyhedral angle < 360°. Art. 583. 1. Each Z of an equilateral triangle is 60°. Art. i.ii, 3 X 60°, 4 X 60° and 5 X 60° are each less than 360°; bat any larger multiple of GO° = or > 360°. .". but three regular polyhedrons am be formed with equilateral &. as faces, 2. Each Z of a si/Krtc* contains 90°. An. irii. 3 X 90° is less than 300°, but any larger multiple of 90" = or > 360". .*. but one regular polyhedron enn be formed with Squares as faces. 3. Each i^ of a regular jiewiai/oM is 108°. Art. \--i. 3 X 108" is less than 300°, but any larger multiple of 108° > 300°. yGoosle Jii)^ HOOK VII.' SOLID riEo:Mi:'niv .■. but one regular piil.\ lu'dmn u;ui lu' formed wHli regu. lar peutagons as faeos, 4. Eaob Z of a regular iu'xaffoii is 120°, anil 3 X 120" .". no regular polyhedron eau be formed with hexagons or with polygoiiB with a greater number of sides as faces, .'. but five regular polyhedrons are 666. T!ie construction of the regular polyhedrons, by ths nse of cardboard, may be effected as folltjws; Draw on a piece of cardboard the diagrams given below. Out the cardboard half through at the dotted lines and eu- tirely through at the full lines. Bring the free edgea together and keep them in their respective positions by some means, snch :is pasHnft- strips of paper over tViem. TTfM P^^ W'i^p -z^ yGoosle POLYHEDRONS ii))iS POLYHEDRONS IK SEHESAL PliOPOf^ITION XXY. TllEOKEM 667. III. inn/ polnhedron , the nitnibi'r of edges increased by two eqiuils the number of vertices increused hy the miinher offacen. Given the polyhedron AT, with the number of its ver- tices, edfjes and faces denoted by V, E and F, respectively. Toprove E + 2=V+F. Proof. Taking the shigle face ABCD, the nuraber of edges equals the number of vertices, or E= V. If another face, CRTT)., be annexed (Fig, 2), three new edges, 6'i?, BT, TI), are added and two new vertices, B and T. :. the mimber of edges gains oue on the aumber of ver- tices, or E^ T+ 1. If still another face, BQRC, beflunexed, two new edges, BQand Qli, are added, and one new vertex, Q. .: E= V+2. With each new face that is annexed, the number of edges gains one on the number of vertices, till but one face is lacking. The last face increases neither the number of edges nor of vertices. Hence number of edges gains one ou number of vertices, for every face except two, the first and the last, or gains F~1 m all. .■. tor thi- entire tiK'n-c, /■:= I'-f- F—'l. That is £.' + 12= t'-i- t\ A.V. 3. Q. s. n. y Google .i'Ji ]!OOK vn . SOLID GEOJIETEY Proposition XXVI. Theorem 668. The sum of Ike fncc amjlex of any j'ohjhedron equals four right ant/Us iaknn us inunij times, less two, (o,- the polyhedron has vtiikfs. Given any polyhedron, with the sura of its face angles denoted by S, and the number of its vertices, edges and faces denoted by V, E, F, respectively. To prove 8= { V—2) 4 rt. A . Proof. Each edge of the polyhedron is the iatersection of two faces, ,■. the number of sides of the faceE = 2 E. :. the sum of the interior and exterior A of the faces = 2 S X 2 rt. A, or EX i rt. A . Avt, 73. But the sura of the exterior A of each faee = 4 rt, A . :. the sum of exterior A of the Ffaces = i'X4 rt. A. Subtracting the sum of the exterior A from the sum" of all the A, the Bujn of the interior A • of the P faces = (.SX4rt. A)- -{FX4 rt. A). Or li={E-F) 4rt. A But E + 2=r+F. Art. 6QT. Hence E~F^V~2. As. 3. Substituting ■,tor E—F, S=(V~~2) 4 r t. A . Ax. 8. Q. I. B. E5, Veri f J the laal two tlieorema iu the uase of the cube. y Google COMPARISON OF POLYHEDKONS COMPARISON OF POLYHEDKONS. SIMILAR POLYEEDKOWS Proposition XXVII. Theoreji 669. // two tetrahedrons have a trihedral angle of one equal to a trihedral angle of the otJter, they are to each other as the products of the edges including the equal trihedral angles. Given the tetrahedrons O-ABC aud O'-A'B'C, with theirvohiines denoted by Faad V, respectively, and having the trihedral A and 0' equal. V OAXOBXOC To prove yr ^yA' X O'B' X O'C' Proof. Apply the tetrahedron O'-A'B'C to (h-ABC so that the trihedral Z (y shall coincide with its equal, the trihedral Z 0. Di-aw CP and O'F 1 plane OAB, and draw OP the projection of OC in the plane OAB. Taking OAB and OA'B' as the bases, and OP and G'P' as the altitudes of the pyramids 0-OAJi and C'-OA'B', respectively. V A OAB X CP A OAB ^ CP But A OA'B' XC'F A OA'B' A OAB OA X OB -Xr 07" Art, <ir>'> OA' X OB' CP _ OC A OA'B' In the similar rt. A OCPand OC'F, T^ OAX OBX OC ^ OA X OB X OC ■'■ V" OA' X OB' X OC O'A' X O'B' X O'C' yGoosle ?,nc, SOLID GEOMETKV 670. Dep. Similar polyhedrons are polyhedmus having the saaie iinmbei' of (aces, similar, each to each, and simi, lai-ly placed, and having thcii- eorrespondiog polyhedral augles equal. PKOi'OSiTio^- XXVni. ThEORG-M 671. Any iiro s-iimlar polyliedrons way he (leeompoaed into the t:ftme niimle.f of lef raited rons, similar, each to each and similarly placed. Given P and P', two similar poiyhfidrons. To prove tlv;it P and P may be decomposed into the same number of tetrahedrons, similar, each to each. Proof. Take R and W any two homologous vertices of P and P'. Draw homologous diagonals in all the faces of P and P except those faces which meet at H and H', sepa- rating the faces into corresponding similar triangles. Through H and each face diagonal thus formed in P, and through S' and each face diagonal in P', pass planes. Each corresponding pair of tetrahedrons thus formed may be proved similar. Thus, in E-ABC and II'-A'B'C, tlie A EP,A and H'B'A' are similar. Art. 329. In like manner A HBC and H'B'C are similar; and &■ ABC and A'B'C are similar. y Google SIMILAR POLYGONS 397 ., NA /HB\ EC /BC\ AC . „. .-. A ARC and A'WC .-u-e simiiar. Art, 326. Hence the corresponding faces of H-ABC&mlH'-A'B'C are similar. Also their homologous trihedral li are equal. Art. 5S4. .". tetrahedron S-ABC is similar to H'-A'B'C. Art. 670, After removing S-ABG from P, and H'-A'B'C from P', the remaining polyhedrons are similar, for their faces are similar, and the remaining polyhedral A are equal. As. 3. By continuing this process, P and /" may be decom- posed into the same number of tftrahedroos, similar, each to each, and similarly placed, Q. E. D. 672. Cor. 1. The hoDwloijoim vdyes of similar polyhe- drons are proportional; Any two homologous line'i in iiro aimilar polyhedrons iiave the same ratio ws any other tiro homologous Hwh. 678. Cor. 2. Any tiro liiiimloijous faces of iiru .similar polyliedrons are to each other «« the nquaret: of any tivo homologous edges or lines; The total areas of any tiro similar polyhedrons are to each other as the sguares of any two homologous edges. Es. 1. Iiithefigure.p.39,i,if tl and those meeting at 0' ave *, 6, : of the tetrahedrons. Kz. 2, If the linear dimeasions of one room are twice as great as the eorrespondiag dimensions of another room, how will their enrf aces (and ,". cost of papering) compare 1 liow will thi'iv volumes compare T Ex. 8, How many 2 iu. cubes can be cut from a 10 in, cube t Ei. 4, If the hasL's of a prisraoid are raetansl^s whose dimensions we 0, b a:id b. a. and altitude is E, And the formula for the volume. y Google 674. each otii BOOii YII. SOLID GEOMETRY PROPOSITION XXIX, Thbokeji Till', volume:; of tiro similar tdmhcdrons r ns the cull's of any pair of Jioniologous Given the similar tetrahedrons O-ABC and O'-A'S'G'. To prove — — -.m:-,' 1" O'A'' ^ . y OAX OBXOG ,^, „, V (yA' X O'B' X O'C ^OA y OB y OG O'A' O'B' O'c' OA ^ OB __ PC O'A' O'B' O'G' V OA yO^y^ OA ^ ~0A^ ' r'~0'A' O'A' O'A' ooT'^ El. 1. In the above figures, if AB=^2 A'B', fiud the ratio of, V to r. Find the same, if AB^U A'B'. Ex. 2 The meaBUremeDt of the volume of a, regular triangular prism reduces to the laeasurBment of the lengths of how many straight lineB ! of a fmstHin of a regular square pyramid ? Ex. 3. Show how to eonstruet out of pasteboard a iBgular pli^m, B puallelapiped, aad a truncated siiua.re prUtn. y Google similar polygon's s Proposition XXS. Theorem 675. The volumes of any two similar polyhedrons i to each other as the cubes of .any two homologous edges, of any other two homologous lines. Given the polyhedrons AK and A'K' having their vol- umes denoted by Fand V, and RB and B'B' any pair of homologous edges. To prove — "^ "•-.:".•:' Proof. Let the polyhedrons be decomposed into tetra- hedrons, similar, each to each, and similarly placed. Art. 671. Denote the volumes of the tetrahedrons in Pby t-i, vs, i-'i . . . and of those in J" by f'l, v'2, v'-i . . . Then V': 1 H'B' Also (/"• each of these rat ''°'~Wb' j) Art. 674, As. 1. " t n + t '1 + «'. + - - ^\' that is, V .^1 Art. 3la, y Google iW HOOK Vil. SOLID GLO^TE'l'in: TWKOitEMS rdXCEHNJNC POLVHEOROXS Ei. 1. TliB Iriteral faees of :i rigbt prism are rectangles. Ex. 2. A diagonal plaue of ti prism is parallei to every lateral edge of the prism not coiitnined in the plnue. Ex. 3. The diagouais of a parallelepiped bisoci^ each other. Ei. 4. The square nf a diagonal o£ a reijtiingular parallelepiped' oqaaJs the sum of the squares of the thrive edges meeting at a vertex Ei. 5, Each lateral face ol :\ prism is pavallel to every lateral edgo not contained in the face. Ex. lateral 6. Every section of a prism made by a plane parallel to a Bilge is a parallelogram. Ex. allel ti prism 7. It any two diagonal planes o' D each other are perpendicular t is a right prism. the base of the prism, the Ex. Iflse is 8. What part of the volume of . a face of the cube and whose vei 3 cube is the pyramid ■fex is the center of the whose onbef Ex. fl. Any section of a regular squ throngh the asis is an isosceles triangle. are pyramid made by a plane Ex. 10. lu any regular tetrahedron, the perpendicular from its foot to any !'; an altitude equals three times Ex. 1 1. Id any regular tetrahedron, , ail altitude equals the i .urn of the perpendiculara to the faces from any point within the tetratedron. Ex. 12. Find the simplest formula for the lateral area of a trun- cated regular prism of n sides. Ex. 13. The sum of the squares of the four diagonals of a paral- lelepiped is equal to the sum of the squares of the twelve edges. [Sno. Use Art. 352.] Ex. 14. A parallelopiped iij symmetrical with respect to what point? Ex. 15. A rectanglar parallelo piped is symmetrical with respect to how many planes? (Let the pupil make a definition of a figure syni- metrical with respect to a plane. See Arts. 4S6, 487.) y Google EXERCISES OX POLYHEDHOXH -JOl Ei. 16. The volumfl of a pyr'amid ivbose lateral odges mn the three edges of the parallelepiped m«eting at a point is what part of the volume ot the parallelopiped 1 Ex. 17. If a plane be passed through a vertex of a cube and tlie tliagonal of a faae oot adjacent to the vertex, what part of the volume of the enbe is eoataioed by the pjrainiil so formed ! Ex. 18. If the angles at tho vertex of a triangular pyramid are right angles and each lateval edge equals a, show that the volume of Es. 19. IIoiv large is a dihedral anglo at the base of a regular pyramid, if tho apotheia ot the base equals the altitude of tho pyramid 1 the lateral surface. Ex. 21. The section of a triaiiguiar pyrainiil by a plane parallel to two opposite edges ia a parallelogram. If the pyramid is regular, what kind of a parallelogram does tha section become ? Ex. 22. The altitude of a regul; ot the base into seguients whii.'h ari :ir tetrahedron divide! Ex. 23. If the edge of a reguh elant height is ^ ; and hence th ir tetrahedron at the altitude '* 3 ' ume is ^^' Ex. 24. If the midpoints o£ all the edges of B- tetrahedron except two opposite edges be joined, a parallelogram is formed, Ex. 25. Straight lines joining the midpoints '^t the opposite edges of a tetrahedron meet in s point and bisect each other. Ex. 26, The midpoints ot tl^- ed-es of a reg' ^e vei'.ic..^ ot a regulur octiihedron. y Google JIOOK Vll, SOLID GEOMKJ'HV EXERCISES. CROUP Ca intOELKMS CONrKRSING POLYHF.DBONfi isect the volume ot a given [insm by a plane parallel U Ex. 3. Through a given point pass a plane which shall bisoct the volume of a given paralleloplpeil. Ex. 4. Given an edge, couatruet a regular tetrahedi'ou. Ex. 5. Given an edge, eonstruct a regular octahedron. Ex. 6. Pass a plane tiirougb the axis of a regular tetrahedron eo that the section shiili bo an iaoaceles triangle. Ex. 7. Pass -J, plane through a cube so that the section nhall be a regular hexagon. Ex. 8. Through three given lines no two o£ which are parallel pass planes which shall form a parallelepiped, Ex. 9. Prom cardboard eonatiuct a regular square pyramid each o£ wbose faces is an equilateral triangle. EXERCISES. GROUP 69 REVIEW F.XEKCISES Make a 1 ist of the pr operties of Ex. 1. Straight li nos in space. Ex. 9, Eight prisms. Ex.2. One line a nd one plane Ex. 10. Parallelopipeds in gen- Ex.3. Two or n lore lines an ,d ^'■'''■ one plane. Ex, 11. Rectangular parallelo- Ex.4. Two plane B and one lin e, pipeds. Ex.5, Two plane sand two line g Es, 12, Pyramids in general. Ex.6, Polyhedrons in general. Ex. 13, Regular pyramids. Ex, 7. Similarpolyhedrons. Ex. 14 Frusta of pyramids. Ex, 8. Prisms in general. Ex, 15, Truncated prisms. y Google Bdox VIIT CYLINDERS AND CONES CYLiHDEKS 876. A cylindrical surface ii5 a carved surface generated bj a straight line which moves so :in constantly to touch a given fixed curve and constantly be pLirLilL-l to a given fixed stniigiit line Thus, every stiadow cast by a pomt of light at a great distance, ao l>y a Btar or the sun, appTOximatPs the oylindrioal form, that is, is, Ijouaded Cjimdrnal surface by a, oylindrieal Hurfaoe of light. Henee, in all radiations (ns of light, heat, magDetism, etc.) from a point at a great distance, we are concerned with oylindrieal snrfacea and solids. 677. The generatrix of a cylindrical surface is the mov- ing straight line; tlie directrix is the given curve, as CDE; an element of the cvlindrical suffice i'; the moving straight line m any one of its positions, as BV 678. A cylinder lo i solid bounded bj ^ cylindrical sniface md L\ two piialiel planes. The bases of a c\lmdei ul its puallel plane faces; the lateral surface is the cylindrical surface iiiclud d liLt\\eeu the parallel planes foiming its bT^f s tlic alti tude of a eyliudHi is the distiin,e lictwteu the 1 The elements of a e^lmhi ait the elements of the cylindrical surface boundm': it. yGoosle iO[ JiOOK Yin, SOLID GEfniKTKY 679. Property of a cylinder inferred immediately. All the elements of a cylinder are equal, lor they ;ire parallel lines included between parallel plai.u's (Arts. 533, 076), Tlie cjiinilers most important in practical lite irt, those determined by their stability, the ease with wliich tlipy can bi, iiude from com- 680. A right cylinder is & cyhndci wliose elements are perpendicular to tlie bases. 681. An oblique cylinder isoue-nliui elements are oblique to the bases, 682. A circular cylinder is a cjlmd i whose bases are circles. 683. A cylinder of revolution is a c\liii- der generated by the revolutiou of a lect angle about one of its sides as an axis. Hence, a cylinder of revolutioa is a right circular cylinder. Some of the properties of tills solid ore derived most readily Tiy cousidering it as generated by a re- volving rectangle ; and otliers, by regnrdiag it an a particular kind of cylinder derived from t!ie grnLial 684. Similar cylinders of revolution ne f\lmdeib gen- erateil by similar ructaiigles revolving ibout homologous sides, 685. A tangent plane to a cyliud \ is t, plane whioh contains one element of the cylinder, and whi-'h do^!^ not cut tlie cylinder on being produced. Es. 1. A plane passing through a taugent to the base of a circu- lar cylinder and the element drawn through the point of contact is tangent to the cylinder. (For it it is not, etc.) Ex. 2. If a plane is tangent to a dicular cylinder, its intersection villi tlic lilane of the base is tangent to the base. y Google CYLINDERS 686 A prism inscribed lu a cylinder is a pr lateral edge'^ ue ebmeiits of the cylinder, and wnos ni I H m tlur- 1 I es of the cylinder. 405 prism whose 687 A prism circumscribed about a cylinder is a prism whose litenl fice-. are tangent to the cylinder, and whoso bases are polygons eii-enmscribed about the bases of the cylinder. 688. A section of a cylinder is the figure formed by the intersection of tlie cylinder by a plane. A right section of a cylinder is a section formed by a plane perpendicular to the elements of the cylinder. 689. Properties of circular cylinders. By Art. 441 the area of a circle is the limit of the area of an inscribed or circumscribed polygon, and the circumference is the limit of the perimeters of these polygons; hence 1. The volume of a circular cylinder is the limit of the volume of an inscribed or circuiAscribed prism. 2. The lateral area of a circular cylinder is the Umil of the lateral area of an inscribed or circumscribed priism. Also, 8. By metJiodf loo advanaedfor this book, iC may be pravfd Ih-il the jierimeUr ol'u rig/it atttuin is the limit of the perimetm- of a rif/hl ner- tvm of an insrvib&l or drcinnMri'ied pri»'n. y Google 690. Ei-enj section of a rylUuli-r mcuh hi) a plane pass- ing tkrou'jh an ekmcid is « parnllrloarain. Given the <>ylindpf AQ cut by a ■plixuc passing through the element AB and fonriiiig the section ABQP. To prove A BQP a CJ . Prooi. API! BQ. Art. 631. It reraaiES to prove that PQ is a straight line II AB. Through P draw a line in the cutting plane ll AB. This line will also lie in the cylindrical surface. Art. 676. .■. this line must eoineide with FQ, (fnr the fine drawn lies in both the cutting phmc and the cylindrical surface, liencc, it iiiiiKt lie their iatvrscction). :. FQ is a straight line || AB. :. ABQP is a /ZJ . (Why?) (J. E. D. 691. Cor. Every section of a rirjht cylinder made, by a plane parsing through an element is a rectangle. Ex. 1. A doov siTiii!;iii;; on its hinges generates what kind of a Es, 2. Every sectioQ of a paBiillelopipeii maJe by a plane iater- secting all its la,teral edges is a imrullciogiaw, y Google CYLINDEBS Proposition If. Theorem The bases of a cylinder are equal. Given the cylinder AQ with tlie bases AVB and CQB. To prove base APB = base GQD. Proof. Let AC and BD be anj' two fixed elements in the snrface of the cylinder AQ. Take P, any point except A and B in the perimeter of the base, and through it draw the element PQ. Draw AB, AP, PB, CD, CQ, QI). Then ACand Bi> are = and II. (Why!) .-. AD is a ^17 . (Why?) Similarly AQ and BQ are 07 . .-. AB^CD, AP^CQ, and P.P=T)Q. (Why?) .■. AAPB=A CQD. (Why?) Apply the base APB to the base CQD so tl.at AM coin- cides with CD. Then P will coincide with ^, (for A AFB = A CQD). But P is any point in the perimeter of the base APB. .: every point in the perimeter of the lower base will coincide with a corresponding point of the perimeter of the upper base. .'. the bases wiil coincide and arc equal- Art. 47. ^, E, B. y Google 408 HOOK Vm. fiOT.lD fJT:0:METl!Y 693. Cob, 1. The sscfioit.'i of a cjlinder madp by two paraJlfl phities cutting all the elements are equal. For the sections thus formed ai'e the bases of the eylinder- iucluded between the cutting planes, 694. Cor. 2. Attn section of a cyHmhr paniUel to the base is equal lo the bat^e. pROPOsiTiON III. Theorem 695. TJie lateral urea of n cirmdnr vylimler is equal fo tlie product of the perimeter of a ritjltt section of the cylin- der by an elemertf. Givea the circular cylinder AJ, haviu" its lateral area denoted by 6', an fiknifnt by JS, and the perimeter of a right section by P. To prove S=PXE. Proof. Let a prism with a regular polygon for its base be inscribed in the cylinder. Denote the lateral area of the inscribed prism by 8', and the perimeter of its rijjht section by P'. Then the lateral edge of the inscribed prism is an ele- ment of the cylinder. Constr y Google 4ft. asa, 3. (Whyf) (Why?) Q. E. D. CYLINDKKS 409 .-. S'^PXE. Art. COS. If the number of lateral faces of the inscribed prism be indefinitely increased, iS" will appi'oaeh ^ as a limit. Art. 689, 2. P' will approach P as a limit. Art. 689, 3. And P'Xfiwill approach PXE as a limit. But. ,S" = PX.E always. .-. 8 = PXE. 696. Cor. 1. The laieml area of a cyliwhr of involu- tion is equal I" the produci of Ihe civionferpnee of its base, by its altHiirh. 697. Formulas for lateral area and total area of a cylin- der of revolution. Denoting the lah-ral area of a cylinder of revolution by S, Ike total area by T. the radius by R, and the altitude by U. T=2 TtRH + -2 nR" :. T^2 nR {H^ R). Ex. 2. If the a of the Lase (H = l terms of li t also, in tenns nf // f Ex. S. What do they liecorae, if the altitude eqwals tlie diametsr <rt the base ? Ex. 4. Ill a cyliniier of revolution, what is the ratio of the latersl "^ea to the area, of the base f to the total area ! y Google 4H) BOOK VTII. SOLID OLOMKTP.V I'HOropmuN i\. TiUiORicM 698. The. vnlums of a riycular cylinder is equal to the product of its base hij Hh tdlitudc. Given the eircnlav I'yli'iLler AJ, having its volume denoted by V", its base by B, and its altitude by K. To prove r=BXM. Proof. Let a prism having a regular polygon for its base be inscribed iu the cylinder, and denote the volume of the inaeribed prism by T', and its base by B'. The prism will have the same altitude, H, as the cylinder. .-. V'^B'XH. (Why?) If the number of lateral faces of the inscribed prism be indeflaitely increased, V will approach V as a limit. Art. 689, i, B' will approach B as a limit. (Whj!) And B'X H will approach BXH asd. Hmit (Wh?)) But V'^B' X ff always. (Why?) .-. V=BX B. (Whys) Q. E. a. 699. Formula for the volume of a circular cylinder. By use of Art. 450, y Google CYLINDERS 411 Proposition V. Theorem 700. The lateral areas, or the total areas, of hco simi- lar cylinderi' of revolution are to each other as the squares of their radii, or as the squares of their altitudes; and their volumes are to each other as the cubes of their radii, or as the cubes of their idfitudcs. Given two similar cyliiidura of revolution having their lateral areas denoted by S and S', their total areas by T and T', their volaraes by T^and P, their radii by R and B', and their altitudes by Sand H', respectively. To prove S -. H'^T: T^BT- : B'^-^E" -. B'^\ and y : y' = E? : R'^ = E'' : S'\ _ H+R U'~R' R' + R'' 2 rcitB BXE _R ^,H ^R" ^g" Proof. Arts, Z'il, 309, S' r I TiR'H' R' X B' T,y~rF- ir- (Wliy?) I 7tR' {W + R') B' ff' + E' B'- (Whyt) (Why f) Q. E. B. Ex. If a eylindrlcal c y Google CONES 701. A conical siirface is a sur- face giaiierated by ii straight iiin' which moves so as constantly to toucli a given fixed curve, and constantly pass through a given fixed point. Tb-tts every shado\r ciist by a ne p of iigM iaeonifliil ic foi'iu, tlmt is, is by a conical sui'faee of liglit. Hen li study ot conical surfaees and solid s a portant from the fant that it eone n a oases of forces radiating from a neai p n 702. The generatrix of a surface is the moving straight AA'; the directrix is the give f d " carve, as ABC; the vertex i tixed |ont as 0; an element is the generatrix in anv o e ot t po tons as BB'. 703. Tiie upper and lower nappes oC a conical snrface are the portions above and belon' the vertex, respeetivelv g,s 0-ABCi\w\ O- A' B'C. Usually it is coavenicat lo limit a conical surface to a single nappe, 704. A cone is a solid bounded by a conicai snrface and a plane cuttmg all the elements . 705. The base of a eone is the face formed by the cutting plane; the lateral surface is the bounding coincal surface; the vertex of the cone is the vertex of the i conical surface; the elements of the cone are the elements of the eonieal surface; cone tlie altitude of a cone is the perpendicolar distance from .the vertex to the plane of tho base. y Google 706. A circular cone is i\ con?, whose base is a circle. The axis of a circalm- cone is the line drawn from the vertex to the center of the base. 707. A right circular cone is a c I one 1 e i is perpendicular to the plane of the 1 An oblique circular cone is a cLri, 1 whose axis is oblique to the base. 708. A cone of revolution is a co e ge e rated by the revolution of a rifjlit t about one of its legs as an axis. Hence a cone of revolution ai 1 circular cone are the same solid. ■> 709. Properties of a cone of revolution inferred im- mediately. 1. The altitude of a con^^ of rrmhiiion U the <uis of the cone. 2. All the elements of a vone ,.f rri'<>]i>/i^n are equal. 710. The slant height of a cone of revolution is any one of its elements. 711. Similar cones of revolution are uones gciienited by similar right triangles revolving about homologous sides. 712. A plane tangent to a cone i^ a pliine which con- tains one element of thu uouu, but wliich does not cut the conical surface on being produced. Ek, 1. A plane passing througii a taageut lo llic ba.se of a circular one and the element drawu through the poiat of coutiict is tnngt-nt to Es, 2. ]f a plane is li the plane of Iho base i-, t; s intersection with y Google 713. A pyramid inscribed m a cone is a pyramid whose liitci i! edges are elements of the coi e and whose base is a polygou in scribed in the base of the none 714. A pyramid c i r c u m scribed about a cone is a pjia [_ _, mid whose lateral faces are tin gent to the cone and whose base is a polygon eireumseribed about the base of the cone. 715. Properties of circular cones. By Art, 441 the area of a circle is the limit of the area of an inscribed, or of a circumscribed polygon, and the circumference is the limit of the perimeters of these polygons; hence 1. The vohmie of a cir<;nlar cone is the limit of the vol- ume of an inscribed or eireumseribed pyramid. 2. The lateral area of a circular cone is the limit of the lateral area of an inscribed or circumscribed pyramid. 716. A frustum of a cone is the \ oi tion of tlie cone iuelndcd between th bi e of the eoiic and a plane parallel tl base. The lower base of the frustum base of the cone, and the upper the frustum is the section made plane parallel to the base of the cone. What must be the altitude and the lateral surface of a frustum of a cone; also the slant height of the frustum ol a cone of revolntion ? y Google CONES 415 Proposition VL Theorem 717. Every section of ti cone made hy a plane passing through its vertex is a triangle. Given tlie oone S-AFBQ with a plane passing through the Tertes S, and making tlie section SPQ. To prove SPQ a triangle. Proof. PQ, the intoi-section of tlie base and the cutting plane, is a straight line. (Whj!) Draw the straight lines SP and SQ. Then 8P and HQ must be in the cutting plane; Art. 498, And be elements of the (lonical surface. Art. 701. ,*. the straight lines SP and SQ are the intersections of the eonieal surface and the euttiug pliuie. .', the section SPQ is a triangle, Art, 81. {/or )I is bomuled bij Ihrc: nhviglit lines). Ea. What liiiid of triaufjie Blade bj a plaxie tiivough the >. y Google BOOK VIII. KOLiD IIEOMETKY Proposition VII. Theorj^m 718. El-cry section of u cirnilar voite made by a plane ^parallel to the buse is a civl''. Given the circular eoue SAB with nph a section made by a plane parallel to the base. To prove apb a circle. Proof. Denote the center of the base by 0, and draw the axis, SO, piercing the plane of the section in o. ' Through SO and any element, SP, of the conical stir- face, pass a plane cutting the plane of the base ia the radius OP, and the plane of the section in op. In like manner, pass a plane through SO ami SB form- ing the intersections OB and oh. :. OrWop, and OB]\ob. (Why?) .-. A SFO and 8B0 are similar to A Spo and Sbo, respectively. Art. 328, ■' OP But OP^OB. (Why J) .'. op = ob. {Why t) .". oph is a circle. (Wlyt) Q. E. ». 719. Cor. The axis of a circular cone passes through the center of every section parallel to the base. y Google CONES 417 Proposition VIII. Theorem 720. The lateral arm of a cone of revolution is equal to half the product of the slant height by the drcumferenee of the base. Given a cone of revoltilion having its taterat ares de- noted by S, its slant lieight by L, and the cireiiinforeuce of ita base by C. To prove S^iCXL. Proof, Let a regular pyramid bo eireuinscribed about the cone. Denote the lateral area of the pyramid by S', aad the perimeter of its base by P. . Then S'^iPXL. Art. 612. H the number of lateral faces of the cireumserlbod pyramid be indefinitely increased, S' will approach S as a limit. Art. 7i5, 2. r wilt approach C as a limit. Art. 44i. And J P X L will approach J CXL as a limit. Art. 253, 2. But S'^i PX L always. (Why?) .-. S^^ CXL. (Whyf) Q. £. n. 721. Formulas for lateral area and total area of a cone of revolution. Denoting the radius oi' the base by R, .S'=i {2 Ttfl X L) :. ti=7TRL. Also T=7iiiL +71^- :. T^TcU {L + B). yGoosle BOOK V!II. SOLID (iEOMETRY PkOPOSITIOX IX. TUEOREll 722. The volume of a circular cone is equal to one-thin of the product of its base by its altitude. Given a circular cone having its volume denoted by V its base by B, and its alfcttude by E. To prove V^iBXH. Proof- Let a pyramid with a regular polygon (or its base be inscribed in the given cone. Denote the volume of the inscribed pyramid by F, and its base by B'. Hence V = ^B'XH. Art. gsi. If the number of lateral faces of the inscribed pyramid be indefinitely increased, V will approach F as a limit, (Why ?) B' will approach B as a limit. (Why ?j And i B' X H will approach J B X S" as a limit. ( Wby i) But V'^i B' X ff always. (Why?) r^iBXH. (Why?) q. £. i>. 723. Formula for the volume of a circular c Ex. 1. If, in a cone of revolution, if =3 and JJ = 4, find S,r and F, Ex. 2. II the altitude of a cone of revolution eijuals the radius of the base, what do the formulas for S, T and V become 1 y Google CONES 419 Proposition X. Theorem 724. The lateral areas, or the total areas, of two simi- lar rones of revolution are to each other as the squares of their radii, or as the squares of their altitudes, or as the squares of their slant heights; and th^ir volumes are to each ether as the cubes of these lines. Given two similar cones of revolution having their lateral areas denoted by S and 8', their total areas by T and P, their volumes by Y and V, their radii by jB and E', their altitudes by H and H', and their slant heights by L and I/, respectively. To prove S -. .^'=^T -. T' = IP: R'- = W' -. H'" = J? -. L'"-; and V : V' = E? : E''' = H' : H'^ = L^ : L'K K V L' + E'' Proof. (Vfhy?) (WhjT) 725. such thf triangle. (Why?) Q. E. D. Dek. An equilateral cone is a eone of nivolution ^. a section through the axis ia au equilateral y Google i2i) BOOS VIII. SOLin GEOJlETRi' Proposition XL Theoeem 726. The lateral area of a frustum of a cone of re.voli tioH is equal to one-half the sum of the circumferences of i bases multiplied iy Us slant height. Given a frustum of a cone of revolution having its lateral area denoted by S, its slant height by L, the rarlii of its bases by B and r, and the circumfereuces of its bases by C and c. Toprove S=i (C+c)XL. Proof. Let the frustum of a regular pyramid be cireum- scribed about the given frustum. Denote the lateral area of the eireumscribed frustum by S', the perimeter of the lower base by P, and the perimeter of the upper base by p. The slant height of the circumscribed frustum is L. Hence S'-J {P + p) X L. Art. 643. Let the pupil complete the proof. (J. E. B. 727. Formula for the lateral area of a frustum of a cone of revolution. S~i {2 TtiJ + 2 tit) L. .-. S^n {U + r)L. 728. Cor. The lateral area of a frustum of a cone of revolution is equal to the product of the circumference of Us Midsection by its slant height. y Google CONES 421 Proposition XII. Theorem 729. The volume of the frustum of a circular cone is equivalent to the volume of three cones, whose common alti- tude is the altitmle of the frustum, and whose bases are the lower hase, the upper base, and a mean proportional between the two iases. Givea a fnistura of a eircalar cone having its volume denoted by V, its altitude by H, the area of its lower base by B, and tiiat of its upper base by b. To prove V = h S {11+ b + Vb X h). Proof. Let the frustum of a pyramid with regular poly- gons for its bases be inscribed in the given fmstum. Denote the volume of the inscribed frustum by V, and the areas of its bases by B' and 6'. .-. V' = h S iB' + 6' + l/ff^xT') , (Why t) If the number of lateral faces of the inscribed frustum be indefinitely increased, V will approach V, B' and V approach B and b respectively, and B' X V approach B X b, as limits. Art. 719. Hence, also, B' + b'+V^B' X b' will approach B -\~ b + V'b X J as a limit. Art. 353. But V'= iH {B + J'+V-^' X b') always. (Whj?) .■. V=iE{B + b + l/fiXb). (Why!) yGoosle i'12 ];00K VIII. SOIJI) GEOMETRY 730. Formula for the volume of the frustmu of a circii, lar cone. F= h if (TtA'- + Tir' + VtiH-'x ni^) . Ex. 2. If a eocieal oil-can is 12 iu. iiielii liow much more tin is required to make it tliaa to make a airailai' oil-pan in. high ! How much more oil wiil it hoid ? Ei. 3, The linear dimeneions of i eonieal funnel are three timt-s tbose of a similar funnel. How mxieh more tin is required to walie the first f How much more liquid will it hold f El. 4. Make a similar oompariaon of iiylindrieai oil-tanks. Of EXERCISES. CROUP 70 THBORBMS COKCERNISQ CYLINDERS AND CONES El. 1. Any section ot a cylinder of revolution through its axis is a rectangle. El. 2. On a cylindrical surface only one straight line can be drawn through a given point, [Sua. For if two straight lines could bo drawn, etc.] Es. 3. The intersection of two planes tangent to a cone is a straight line through the rertex. Ex. 4. If two planes are tangent to a cylinder, their line of inter- seetion is parallel to an element ot the eyliuder, [SuG. Pass a plane -L to the elements of the cylinder.] Ei. 5. If tangent planes be passed through two diametrically opposite elements of a circular cone, these planes intersect iu a straight line through the Tertes and parallel to the plane of the base, Ex. 6. In a cylinder of revolution the diameter of whose base equals the altitu<3e, the voiumc equals one-third the product of the total surface hy the radius of the base. y Google EXERCISES ON THE CYLINDER AND CONE 423 Ek. 7. Aeylinder and a cone o! revolutiou have the same base Bnd the same altitude. Ficd tbe ratio of their lateral surfaoes, and also of their voluroes. Ex. 8. It an equilateral triangle whose Bide U a be revolved iiljout ooe of ita Bides as an asis, find (he area generated in terms of a. Ex. 9. If a rectangle whose aidea are a and b be revolved first about the aide a as an axis, and then about the side b, find the ratio of the lateral areas generated, and also of the volumes. Ex. 10. The baaoe of a cylinder ooacentric. The two solids have th ot the base of the cone is twice the < def. What kind of Hue is the iute and how far is it from the base 1 it a cone of revolution are attitude, and the diameter er of the base of the cjlin- a of their lateral surfaoea, Ejc. 11. Determiue the same when the radius of tha «■ imea the radius of the cylinder. Also when r times, Ex. 12. Obtain a formula in terms o£ r for the volume of the 'rustum of an equilateral cone, in which the radius of the upper base a r and that of the lower base is 3r. Ex. 13. A regular hoxagon whose aide oual through the center as axis. Find, i and volume generated. folvea about a diag- Ex. 14. Find the locus of a point at a giv en distance f straight line. Ex. 15. Find the locus of a point whos e distance f line is in a given ratio to ita distance froc a a fixed pis dioular to the line. Ex. 16. Find the lo us of all straight 1 nea which m angle with a given line at a given point. Ex. 17. Find the lo cus of all straight 1 nea which m angle with a given plan e at a given point. Ex. 1 8. Find the locus of all points at c tanee from the surface of a given cylinder of n the surface of a gvv y Google JK vni. SOLID GEOJIIM'KY EXERCISES. QKOUP 71 )N'Cf.KN'IXG Till'; l'VLlSr>F.[l AND CONE Ex. !. Through a given element o( a tireular eyliader, paaa v plane l.augent to tbe ejlicder. Ex. 2. Through a given element of a circular ciine, pass a plane tniij^Li Ex. 3. About a given circular oylind s regular polygon for it^ base. Ei. i. Through a giren point outside a circuUr cylimier, paas a plane tangent to the cyiinder. Ex. B. Through a given point outside a given elnmlar eone, pas3 a plane tangent to tbe cone. [St'ti. Through tbe vertei of tlio cone and the given point pass a line, and produce it to meet the plane of the base.] Ex. 6. Into what eegiuents must the altitude of a coneof revoliitirjn he divided by a plane parallel to the base, in order that the volume of the cone be bisected? Ex. 7. Divide the lateral surface of a given cone of revolution into two equivalent parts by a plane parallel to the base. Ex. 8. If the lateral surface of a cylinder ot revolution be ont along one element and unrolled, what Movt of a plane figure is formed f lience, out of cardboard eonatruot a cylinder of revolution with given altitude and given eireumterenee. Ex. 9. If the literal ^ uitice of ac one of revolution be out along one element and umoHed what sort ol : a. plai 16 figure is formed ! Hence, out ol taidboa li toi JStiuet a com e of revohition ot given slant height Ex. 10. C(n= trmtan e^nlla teril CI TOO ont of pasteboard. Ex. 11, eout itruit a 1 luatliu . Of a< ■one of revolution out , of paste- board. y Google Book IS THE SPHERE 731. A sphere is a solid bounded by a surface all points of which are equally distant from a point withiu called the center. 732. A sphere may also be defined as a solid generated by the revolution of a semicircle about its diameter as an axis. Some oE the properties of a epheio may be obtained more readily from one o( the two deflcitions given, and some from the otliec. A sphere is named by naming tiie point at its eecter, or by naming three or more points on its surfaae. 733. A raduis of a sphere is a line drawn from the center to any point on the surface, A diameter of a sphere is a line drawn through the center and terminated at each end by the surface of the sphere. 734. A line tangent to a sphere is a line having bnt one point in common with the surface of the sphere, however Jar the line be produced. (425) yGoosle 4::f) BOOK IX. SOLID GEOJffiTRY 735. A plane tangent to a sphere is ;i plane having bu one point in common with tlifi surface of the sphere, how- ever far the plane be produced. 736. Two spheres tangent to each other are spheiv-- whose surfaces have one poiut, and only one, in common. 737. Properties of a sphere inferred immediately. 1. All radii of a sphere, or of equal spheres, are equal. 2. All diameters of a sphere, or of equal spheres, are. equal. 3. Two spheres are equal if their radii or their diame- ters are equal. PROi'OSiTioN I. Theorem 738. A section of a sphere, tnade by a plane is a circle. Given HI 1 1 1.) (i I id PCD a section made by a plane cutting the sphere. To prove that FGD is a circle. Proof. From the center 0, draw OA X the plane of the section. y Google THE SPHERE 427 Let Che B. fixed point on the perimeter of the section, and P any other point on this perimeter. Drawee, AP, OC, OP. Then the A OAP and OAG hvb rt. A. Art. 505. OP=OG. (Whj!) OA^OA. (Why!) A OAP=A OAC. (Whjf) .-. AP=AC. (Wiiy t) But P is any poiut on the perimeter of the section PCD. :. every point on this perimeter is at the distance AO from A. .". PCD is a circle with center A. Art. 197. q. E. s. 739. Cor. 1. Circles which are se.ctions of a sphfre wade by planes equidistant from the center are equal; and conversely. 740. CoE. 2. Of two circles on a sphere, the one nuide by a plane more remote from the center is smaller; and conversely. 741. Dep. a great circle of a sphere is a circle whose plane passes through the center of the sphere. 742. Dep. A small circle of a sphere is a circle whose piane does not pass through the center of the sphere. 743. Dep. The axis of a circle of a sphere is the diameter of a sphere which is perpendicular to the plane of the circle. Thus, on figure p, 42fi, BB' is the axis of PCD. 744. Dep. The poles of a circle of a sphere are the "istremities of the axis of the circle. Thus, P and B', of ■igure p, 426, are poles of the circle PCD. y Google 428 BOOK i\. SOLID f;EOMETKY 745. Properties of circles of a sphere inferred imiiif diately. 1. 2'Ae axiK of a circle of a sphere pnxses through tl center of the circle; and conversely. 2. Parallel circles have the same axis and the same poln 3. All great circles of a sphere are equal. 4. Every great circle on a sphere bisects the sphere ai, its surface. 5. Two great circles on a sphere bisect each other. For the line of intersection of the two planes of th circles passes througli the center, and hence is a diametf' of each circle. 6. Through two points (not tite extremities of a diameio on the surface of a sphere, one, and only one, great circle ch he passed. For the plane of the great circle nmat also pass throng the center of the sphere (Art. 741), and through thro points not in a straight line only one plane can be passe (Art. 500) . 7. Through any three points on the surface of a sphen not )rt the same plane icith the center, one small circle, a/t only one, can he passed. 746. Def. The distance between two points on the sui face of a sphere is the length of the minor arc of a grt-a circle joining the points. Ex. 1. If the raiii.is of a sphere is 13 in-, find the vadlua c circle on the sphere made by & plane at a distance of 1 ft. from center. Ex.2. Wliat geographical eirclea 00 the earth's surface are gn and what small circles t Ex. 3. What is the largest nuiabt ir of points in which two cir ou tlie surface of a aphevo can iutersi Eictf Why? y Google THE SPIIEKE 41!y Proi'OSition II. Theorem 747. All points in the circumference of a circle of a sphere are eqvnlhi lU^tnut from each pole of the circle. Given ABC a circle of a sphere, and P atid P' its poles. To prove the arcs FA, PB, PC equal, and ares F'A, r'B, P'G equal. Proof. Draw the chords PA, PB, PC. The chords PA, PB and PC are equal. Art. 518. ,■. area FA, PB and PC are equal. Art. 313. la like manner, the arcs F'A, F B and FC may be Q. E. B. proved equal. 748. Def. The polar distance of a small circle on a sphere is the distance of any point on the circumference of the circle from the nearer pole. The polar distance of a great circle on a sphere is the dis- tance of any point on the eircumfereuce of the great circle from either pole. 749. Coil. The polar distance of a great circle is the quadrant of a great circle. y Google 430 BOOK IS. SOLID GE0\£TE¥ Proposition III. Theoreji 750. // a point on the surface of a sphere is at a quaii rfinf's dhhmce from two other poinl.i on the. surface, it i. the pole of the great circle through those points. Given PB and PC quadrants on the surface of the spliere 0, and ABC a, great circle through B and C. To prove that P is the pole of ABC. Proof. From the center draw the radii OB, OC, OP. The arcs PB and PC are quadrants. (Why t) .■- ^ POB and POO are rt, A. (WhyT) .-. PO ± plane ABC. (Why?) .". P is the pole of the great circle ABC. (Why i) Q. 1. D, 751. Cor, Through two given points on the surface of a sphere to describe a great circle. Let A and B be the given points. From A and B as centers, with a quadrant as radius, describe arcs on the surface of the sphere inter- secting at P. With P as a center »nd a quadrant as a radius, describe a great circle. y Google THE SPHEliE Proposition IV. Theorem 752. A plane perpendimlar to a radius at its extre is tangent to the sphere. Given the sphere 0, and the place MN J. the radius OA of the sphere at its extremity A. To prove MN tangent to the sphere. Proof. Take P any point in plane MN except A. Draw OP. Then OP>OA. (Whj-f) .■. the point P is outside the surface of the sphere. But P is any point in the plane MN except A. .: plane MN is tangent to the sphere at the point A, (for every point in Vie plane, except A, U milstde the surface fif t!ie sphere). Art. 73&. a, B. D. 753. Cor. 1. A pkinc, or a line, which is tangent to a Sphere, is perpendicular to the radius drawn to the point of <:ontact. Also, if a plane is tangent to a sphere, a perpendicii- far to the plane at its point of contact passes through the center of the sphere. 754. Cor. 2. A straight line perpendicular to a radius ''J a sphere at its extremity is tangent to the i^phere. 755. Cob. 3, A straight line tangent to a circle of a Sphere, lies in the plane tangent to the sphere at the point of y Google 4:62 BOOK IX. SOLID GKo:\iT:T]iy 756. Cob. 4. A straight line drawn in a tangent plan and through the point of contact is tangent to the sphere : that point, 757. Cor. 5. Two straight lines tangent to a sphere , a given point determine the tangent plane at that point. 758. Def. a sphere circumscribed about a poiyhedro is a sphere in whose surface lie all the vertices of tli iiolyhedron. 759. Df.p. a sphere inscribed in a polyhedron ij; sphere to which all the faces of tlic polyhedron are tangent PROPOSITION V. Problem 760. To circumscribe a sphere about a give hedron. Given the tetrahedron ABCD. To circumscribe a sphere about ABCD. Construction and Proof. Construet E and F the centers of circles circumscribed about the A ABC and BGD, re- spectively. Art. SSfi. Draw ES X plane ABC and FK X plane BCD. Act. 5U. Draw FG and FO to G the midpoint of BG. yGoosle THE SPHERE Add Then EG and FG are X BC. Art. ii3. .-. plane EOF 1 BC. Art. 509. .-. pljine EOF ± plane ABC. Art, 555. .-. SH lies in the plane FOE. Art. 558. In like manner FK lies in the plane F6E. The lines EG and FG are not ll, [for Ihfy nieel in the point G), :. the lines Bff and -F£: are not !1. Art. 122. Hence ES innst meet FK in some point 0. But EJ/^ is the locus of all points equidistant from A, B and C\ and F^ is the loeus of all points equidistant from B, C and Z>. Art, 520. ,'. 0, which is in both EH and FK, is equidistant from A, B, Oand D. (Whyl) Hence a spherical surface constructed with as a center and OA as a radius will pass through A, B, G and D, anil form the sphere required, q, e. f. 761. Cor. 1. Four points not m Ike same plane deter- mine a sphere. 762. Cor. 2. The four perpendiculars erected at the '"'enters of the faces of a tetrahedron meet hi a point. 763. Cor. 3. The sije planes perpendicular to the edges of a tetrahedron at their midpoints intersect in a point. 764. Dep, An angle formed by two curves is the angle formed by a tangent to eaoh curve at the point of inter- section. 765. Def. a spherical angle is an angle formed \iy two interseetiutr avns of great circles on a sphere, and l>euee by tangents to these arcs at the point of intersection. y Google 34 BOOK IX. POLID r.EOMKTRY PkOPOSITION VI. PltOHLEM 766. To inscribe a sphere in a given ieffihrilrtm. Given the tetraheilion ABOD To inscribe a spheie m ABC1> Coaatruction and Proof. BibPct the dihoilral angle D- AB-C hy the plane OAB; similarly bisect tlie diliedral A ■whose edges are BG and AG hy the planes OBC and OAG, respectively. Denote the point in whieh the three bisecting planes intersect by 0. Every point in the plane OAB is equidistant from the faces DAB and GAB. Art. 562. Similarly, every point in OBC is ec|uidistant from the two faces intersecting in BG, and every point in OAG is equidistant from the two faces intersecting in AG. .: is equidistant from all four faces of the tetra- hedron. Ax. 1. Hence, from as a center, with the J. from to any one face as a radius, describe a sphere. This sphere will te tangent to the four faces of the tetrahedron and .'. inscribed in the tetrahedron. Art. 759. 767. OoR. The planes bisecting the s of a tetrahedron meet in one point. : dihedral angles y Google THE SPHERE Proposition VII. Problem 768. To find the radius of a given material sphere. Given the material sphere 0. To construct the radius of the sphere. Constnictioa. With any point P (Fig, 1) of the surface of the sphere as a pole, describe any" convenient circum- ference on the surface. On this eireumferenee take any three points A , B and C. Construct the A ABC (Fig. 2) having as sides the three chords AB, BC, AC, obtained from Fig. 1, by use of the compasses. Art. 283. Circumscribe a circle about the A ABC. Art. 286. Let KB be the radius of this circle. Construct (Fig. 3) the right A kpb, having for hypot- enuse the chord pb {Fig. 1) and the base ii. Art. 384. Draw bp' ± hp and meeting pk produced at p'. Bisect /jp' at 0. Then op is the radius of the given sphere. Proof- Let the pupil supply the proof. Q. E. F. y Google 436 BOOK IX. SOLID GEOTilKTRY Proposition VI]!. Theorem 769. The intersection of itvo spherical surfaces is the circumference of a circle whose plane is perpmdicularHo the line joining the centers of the spheres, and whose center is i)i that line. Given two intersecting © and 0' wtiinh, by rotation about tlie line 00' as an axis, generate two intersecting splierical siirfaces. To prove that tlie intersection of the spherical surfaces is a ©, whose plane ± 00', and whose center lies in 00'. Proof, Let the two circles intersect in the points P and Q, and draw fhe common chord PQ. Then, as the two given ® rotate about 00' as an axis, the point P will generate the line of intersection of the two spherical surfaces that are formed. But PR is constantly 1 OO". Art. 241, .', PB generates a plane J. 00' Art, 510, Also PR remains constant in length. ,', P describes a circumference in that plane. Art. 197. Hence the intersection of two spherical surfaces is a O , whose plane X the line of centers, and whoso center is in the line of centers. Q. E. B. Tlie abovft demonstration is an illustration of the use of the second definition of a sphere (Art. 732). y Google THE SPHEKE 437 Peoposition IX. Theorem 770. A spherical angle is measured by the arc of a great circle described from the vertex of the angle as a pole, and included between its sides, produced, if necessary . Given /.BAG a spherical angle formed by the intersec- tion of the arcs of the great circles BA and €A, and BO an are of a great circle whose pole is A. To prove /. BAG measured by are BG. Proof, Draw AI> tangent to AB, and AF tangent to AC. Also draw the radii OB and OC. Then Al> X AG. Art. 230. Also OB L AO iJorABisaqm-iranl). :. OB WAD, (Why?) Similarly OGWAF. :. ^B0(J= IDAW But Z BOG is measured by arc BG. :. IDAF, that is, LBAG, is measured by t Art. 538. Art. 257. ■■C BG- (Why?) Q. E. 9. 771. Cor. A spherical angle is equal in the plain ofthv dihtdral angle formed hij (he planer of Us sides y Google 4r,S BOOK IX. HOLIU liEOJIETUY SPHERICAL TRIANGLES AND POLYGONS 772. A spherical polygon is a portion of the a sphere boumlyd by three or more arcs of gi'eat circles, as ABGD. The sides of the spherical polygon are the bounding arcs; the vertices are the points in which the sides in- tersect; the angles are the spherical angles formed by the sides. 773, A spherical triangle is a. splieriea! poiygon of three sides. Spherical triangles are classified in the same way as plane triangles; viz., as isosceles, equilateral, scalene, right, obtuse and acute, 774. Relation of spherical polygons to polyhedral angles. If radii be drawn from the center of a sphere to the ver- tices of a spherical polygon on its surface (as OA, OB, etc., in the above figure), a polyhedral angle is formed at 0, which has an important relation to the spherical poly- gon ABGD Bach face angle of the polyhedral angle equals {in yium- her of degrees contained) the corresponding side of the spheri- cal polygon; Dach dihedral angle of the polyhedral angle equals the corresponding angle of ike spherical polygon. Hence, corresponding to each property of a polyhedral angle, there exists a property of a spherical polygon, and conversely . y Google THE SPHERE 4J9 Ilenoe, also, a trihedral angle and its parts correspond to a spkfricnl triangle and its parts. Of the common proportieB of a polyheilral angle and a epherical polygon, some are discovered more readily from the one figure and some from the other. In general, the spherical polygon ia simpler to deal with than a polyhedral angle. For instance, if a triheUral angle were drawn with the plane angles of its dihedral angles, nine lines would be used, forming a complicated tigure in solid apace; whereas, the same magnitudes are represented in a apherieal triangle by three lines in an approximately plane figure. On the other hand, the spherieal polygon, beeauao of its lack of detailed parts, is often not so suggestive of properties as the poly- hedral angle. Proposition X. Theorem 775. The sum of tu'o sides of a spherical triangle is greater than the third side. Givea the spherical triangle ABC, of whieh uo side is larger tliaii AB. To prove AC-^ BO AB. Proof. From the center of tlie sphere, 0, draw the radii OA, OB, OC. Then, in the trihedral angle O-AHG, /LAOC. + IBOV > AOB. Art. 582. .-. AV-^ BO > AH. Art. 774. Q. E. 5. y Google 440 BOOX IX. SOLID CKOUETllY 776. CoK. 1. Any side of a spherical iriautjleis gfealn- thin the. difference betiveen the other two mles. 777. Cor. 2. TJie shortest path ietween two points on the surface of a sphere is the arc of the great circle joining those points. For any other path between" the two points may be made the limit of a series of arcs of great oircies connect- ing successive points on the path, and the sum of this series of arcs of great circles connecting the two points is greater than the single arc of a great circle connecting them. Proposition XI, Theorem 778. The smn of the siiits of a spherical polygon is less than 360°. Given the spherical polygon A BCD. To prove the sum of the sides of ABCD < 360°. Proof. From 0, the center of the sphere, draw the radii OA, OB, OC, 01). Then lAOB-\- IBOG+ lOOI>-\- IDOA < SCO". (Why T) .-. AB -^ B(] + CD -^ DA <360°. Art, 77*. Q. E. I>. y Google SPHERICAL TKIANGLES 441 779. Def. The polar triangle of a given triangle is the triangle formed by taking the vertices of the given triangle as poles, and deserihing arcs of great cir- cles. (Hence, if each pole be regarded as a center, the radius used iu describing each arc is a quadrant.) Thus A'B'C'is the polar ti'iangle of ABC; also B'E'F' is the poiar triangle of BEF. Proposition XII. Theorem 780. If one spherical triangle is the polar of another, then the second triangle is the polar of the first. Given A'B'C the polar triangle of ABC. To prove ABC the polar triangle of A'B'C . Proof. B is the pole of the arc A'C . :. arc A'B is a quadrant. Also C is the pole of the arc A'B'. .'. arc A'C is a quadrant. .*. A' is at a quadrant's distance from both B and C. .*. A' is the pole of the arc BC. Art. 750. Iq like manner it may be shown that B' is the pole of AC, and C the pole of AB. g. E, 9. Art. 779. (Why t) (Why?) (Why?) yGoosle 442 I'.OOK !X. SOLID aEOlIETUY Phoposition XIII. Theorem 781. Ill ii sjihfrical triangle <iiid its polar, each angle of one Irianglc is the siipphmeiii of the side opposite in the other li-iangle. Art. tso. (Whyf) (Why!) Given the polar ^ ABC ami A'B'C with the sides of ^£(7 denoted by a, b, c, and the sides of A'B'C denoted by «', h', c', respectively. To prove A + «' = 180°, B + 6'=180°, C' + c'=180°, A' + a= 180°, B' + b= 180°, C + c - 180°. Proof. Produce the sides AB and AG till they meet B'C in the points D and F, respectively. Then B' is the pole of AF :. arc B'F=90' Also C is the pole of AD :. arc CD -90' Adding, B'F + C'I>=180°. Or B'F + FC + .BF^ 180°. Ax. 6- Or 7J'C' + 7H'-180°. But B'C'==a', and /ii'' is the measnre of the A A. Art, 770. .-. A + a' -180°. In like manner the other supplemental relations may be proved as specified. Q. a. B. 782. Def. Supplemental triangles are two spherical triangles each of which is the polar triangle of the other. This new name for two polar triangles is due to the property proved in Art. 781. y Google SPHERICAL TRIANGLES 443 Proposition XIV, Theorem 783. The sum of the angles of a spherical triangle is greater than 180°, and less than 540°. Given tliG spherical triangle ABC. To prove J + S + C > 180" and < S40°. Proof. Draw A'B'C, tlie polar triangle of ABO, and denote its sides by a', S', c'. Tiien A + a' - 180" 1 B -\- I' = 180" > Art. m. C + c' = 180" ' .-. A -r B + C + a' + b' + c' = 540°. . . (1) Ax, 2. Hut ^a' + b' + c' < 3ilO' Ai-i. -7R. j „' 4- i' + c' > 0= Subtracting eacb of these In tnni from (1), A + B + C > 180° and < 540". ak. ii. (I. E. B. 784. COH. A spherical triangle may have one, two or three right angles; or it may have one, two or three obtuse angles. 785. Def. a birectangular spherical triangle is a splier- wal triangle containing two right angles, 786. Dei\ a trirectangular spherical triangle is a spherical triangle containing three right angles. y Google 444 HOOK IX. SOLID GEOMETIiV 787. GoR. The surface of it sphere may he dimded into fight inrectaiigiilar spherical triangles. For let three planes X to each other be passed through the center of a sphere, etc. 788. Def. The spherical excess of a sphei-lciil t.ria is the excess. of the sum of its angles over 180°. igle 789. whieh Y Dep. Symmetrical spherical triangles are triangles ive their parts eqnai, but arranged in reverse order. Three planes passing through the isenter of a sphere form a pair of symmetrical spherical triangles on opposite sides of the sphere (see Art. 580), as & ABC and A'Ji'C of Fig. 1. 790. Equivalence of symmetrical spherical triangles. Two plane triangles which have their parts equal, but ar- ranged inreverse p' order, may be a^ /\ made to coincide / \v /^ \ by lifting up one / x. ^ ^ 'X X ^-' triangle, turning it over in space, and plaeiug it upon the other triangle. But two symmetrical spherical triangles cannot be made to coincide in this way, because of the curvature of a spherical surface. Hetiee the equivalence of two sym- metrical spherical triangles must be demonstrated in some indirect way. y Google SPHERICAL TEIANGLES 443 791. Property of symmetrical spherical triangles. Two isosceles symmetrical spherical triangles are equal, for they can be made to coincide. Proposition XV. Theorem 792. Two symmetrical spherical triangles are equivalent. Given the i formed by plane (See Art. 789.) To prove Proof. Let the points A, B, Also draw great ®. Abo P' lymmetrical spherical A ABC and .I'R'C, s passing through 0, the center of a sphere. A AnCOA A'B-C. P be the polo of a small circle passing through Draw the diameter POP'. PA, PB, PC, P'A', P'B', P'C, all arcs of PA = PB t= PC. --PA, P'B' ^ PB, P'C = PC. Ax. 1 Similarly And Adding In case the A'B'C, V-X the , ,'. P'A' = P'B' =P'C'. J and P'A'B' are symmetrical isosceles A PAG^^FA'C. A PBC=^ PB'C, l^PAB-+APAC+APBC ^AP'A'B'+/^FA'C'-\-AP'B'C'. kx. AABC^AA'B'C. Ax. polos /'and 7" fall outside the ^ ABC iiii nuni! simply the demoiistratiou. q. e. b. y Google BOOK IX. SOLID r.EOMK'niY i'Hiii'o^rnoN XVI. Tii 793. On the same sphe? {ingles are equal, I. If two sides and the included angle of one are equal to liro .sides atid the included angle of the oilier; or JI. Ifiivo angles and the inehtded side of one a io two angles and the included side of the other, the corresponding I'quil parts being arranged in the saw/; order in each case. r on equal spheres, ivn ;»■;. ire equitl I. Given the spheriRai A ABC and J>EF, in which AC=i>F, CR = FE, and ZC=IF. To prove A ABC= A DEF. Proof. Let the pupil supply the proof (see Book J, Prop. YD- II. Given the spherical & ABC and DEF, in whieh Z(";=ZF, lJi = /.E, unA CB^FE. To prove A ABG^ A J)EF. Proof. Let the pupil supply the proof (see Book I, Ex. 1. tre 12 in. If the and 3 Ex. 2. 15 inciiea The t£ In dmm lineof centers of two spl: ill,, liow ai'e tho spheres auk on a motor car is a crylltider l!6 inches long and leter. How many gallons of gasolene will it hold 1 equilateral cone, find the ratio of the lateral area to y Google SPHEEICAL TRIASGLES Peoposition XVII. Theorem 794. On the same sphere, or on equal spheres, two tri- angles are symmetrical and equivalent, I. If two sides and the inchtded angle of one are equal to two sides and tlie included angle of the other; or II. If two angles and the inrAuded side of one are equal to two angles and the included side of the other, the cotfesponding equal parts being arranged in reverse order. I. Given the spherical & ABC and DSF, in which AB = DE, AC = DF, iind £A = £D, the eorrespomling parts being arranged in reverse order. To prove A ABC symmetrical with A DEF. Proof. Construct the /MVE'F' symmetrical with ADEF. Then A ABC may be made to coincide with A D'E'F', Art, 703. {haring two sides and the hiclmh-d L etiuat ami arraiigeil in llie >ameor,hy). But A D'E'F' is symmetrical with the A DEE. :. A ABC, which coincides with A D'E'F, is symmet- rical with A hEF. ir. The second yiirt of the theorem same way. ; proved in t! I), E. D. y Google BOOK IX. SOLID GEOMKTKV Proposition XYIII. Theorem 795. If two triangles oti Ike same sphere, or equal spheres, are mnhtally equilateral, they are also imitnally equiangular, and therefore equal or symmetrical. Given two mutually equilateral spherical A ABC and A'B'C oil the same or on equal spheres. To prove A ABC and A'B'C equal or symmetrical. Proof. Prom and 0', the centers of the spheres to ■which the given triangles belong, draw the radii OA, OC, O'A', O'B', O'C. Then the face A at — corresponding face A at (y. Art. 774. Hence dihedral A at = corresponding dihedral A at 0', Art. 584 ,'. A of spherical A .4 £(7= homologous A of spherical /\ A'B'C. Art. 774, .". the A ABC and A'B'C are equal or symmetrical ficcording as their homologous parts are arranged in the same or in reverse order. Art. 789. 796. Note. Theconditions in Props. XVlaad SVIII which make two sphei'iea! triangles equal aro the same as those which malse two plane triangles equal Hence many other propOBitions occur in Bpherieal geometry which are identical with correspondlag proposi- tions in plane geometry. Thus, many of the construction problems of epherioal geometry are Bolvd in. the sama way as the corresponding construction problems m plane jfeomelryi as, to bisect a given aogle, etc. y Google 'SPHEBICAL TEIANGLES 449 Proposition XIX. Theorem 797. If two Iriangles on the same sphere are mutually equiangular, they ore idno mutually equilateral, and there- fore equal or symmetrical. Given the muhially equiangular spherical ii Q and Q' on the same sphere or on equal spheres. iitiiaOy equilateral, and To prove that Q and Q" are i therefore equal or symmetrioal. Proof. Oc.nHtruct P and P' the polar & of Q and Q', respective b . Then A P aud P' are mutually equilateral. Art. 7S1. .". iSv P and P' are mutually equiangular. Art. 795. But Q is the polar A of P, and Q' of £". Art. 780. .". A Q and Q' are mutually equilateral. Art. 78i. Henee Q and Q' are equal or symmetrical, according as their homologous' parts are arranged iu the same or in :.'everse order. Art. 789. Q. E. B. 798. CO!E. If tu-o mutualhj equiangular tiianijtes are 0)1 unequal uplfren, their correspomUng sides have the same ratio as the radii of their respective spheres. y Google 450 liOOK [X. SOLID CEOMETliV I'HOFOSITION XX. TlIEOlJEM 799. Jii an isosceles spherical triangle the angles opp,, site the eqnul sides are equal. Given the spherit-a! A ABC in whii^li An=-AC. To prove Z7J=ZC. Proof, Draw an arc from the vertex .1 to i>, the mid- point of the base. Let the pupil suppb' the roraaiiider of the proof. I'KorosiTiON XXI. Theorem (Conv. of Prop. XX) 800. Jftivo angles of a spherical triani/le are equal, the sides opposite these angles are equal, and the triangle is isosceles. Given the spherical A ABG in which ZB= Z 0. To prove AB^AC. Proof. Construct A A'B'G' the polar A of ABC. Then A'C' = A'B'. Art .-. ^6"= Z7J'. Art .-. AB = AG. Art yGoosle SI'IIEEICAL TRIANGLES 451 Proposition XXII. Theorem 801. 2m any spkt^rical triangle, if two angles are tm- equaJ, the sides opposite these angles are unequal, and the greater side is opposite the greater angle, and Conversely. Given the spherical A ABC in which ,^. JiAO is greater To prove liC > BA. Proof. Draw the iirc AD making IPAO pqw.\\ lo 10. Theu JiA = DC. Art,. 800. To each of these equals add the are HI). :. BD+ J)A^TiT>-\- T>(\ or liC. (Why?) But ill A BBA, BI) + T)A > BA. (Wl.yT) .-. BO > HA. Ax. 8, Let ihe popii ijrnve the eonver.se by the indirect method see Art. 106). Q, £, D. Ex. 1. Bisect a givi^n spherical aHgle. Es. 2. Bisect B given are of a great circle oii a sphei'f- (■'lii^il to a given spherical angla on tlie same sphere. Ex. 4. Fimi (lie ioeiia ot the ccoterB o( tbe oirolea o£ a epher foi'iiifd by pianos perpendloulac to a given ilinmeter of tlip give 'pheru. y Google i'y2 ROLTD GEOMElliY SPHERICAL AREAS 802. Units of spherical surface. j\ may be measured in tonus of, eirlicr 1. The customary units of arm, as square foot, etc., or 2. Spherical degrees, or spherids. 803. A spherical degree, or spherid, is one-ninetieth part of one of the eight trireetangnlar triaugles into which the surface of a sphei-o may be divided (Art. 787), or -^is part of the surface of the entire sphe A solid degree i spherical surface a square inch, a ie-nicet!eth part of a tri rectangular angle (se lU). I sphere 804. A lune is a portion of the surface of i bounded by two semicii-eumferences of great circles, as PBPC of Fig. 1. The angle of a lune is the angle formed by the semieircumfevences which bound it, as the angle BPG. 805. A zone is the portion of the surface of the sphere bounded by two parallel planes. A zone may also be defined as the sur- face generated by an arc of a revolving semicircumfetenpe. Thus, if QFQ' {FIb- 2) generates a sphere by rotating about QQ", its diameter, any arc of QFQ', as ISF, generates a zone. 806. A zone of one base is a zone one of whose boiitiding planes is tangent to the sphere, as the zone generated by the arc QtJ of Pig. 2. pig. -.-. 807. The altitude of a zone is the perpendiuulat- tiis tance between the bounding planes of the zone. The bases of a zone are the circumferences of the circles of the sphere formed by the bounding planes of the zone. y Google Spherical areas 4yd Proposition XXIII. Theorem 808. The area generated hy a straiglit line revolving about CHi axis in its plane is equal to the projection of the line upon the axis, mtiUipUed by tM circumference of a circle whose radius is the perpendicular erected at the inidpoint of the line and terminated by the axis. Given AB and X¥ in tlie same plane, CD the projection of AB on XY. PQ the ± bisector of AB; and a surface generated by the revolution of AB about XY, denoted as "area ATi." To prove area AB^ CD X 2 nPQ. Proof. 1. In genera!, the surface generated by AB is tlie surface of a frustum of a cone (Fig. 1). ,-, area AB=AB X 2 tiFR. am. 72S. Dvaw AFLBD, then A ABF and PQE are similar. Art. 32s. .-. AB: AF^PQ: PR. (Why!) .-. AB X P7^ = AZ'■'X PQ, or 071 X PQ. (Why!) Substituting, area AB= CBX1 nPQ. Ak. 8. 2. If ABWXY {Pig. 2), the surface generated by AB is the lateral surface of a cylinder. .-. area AB- Ci> X 2 tiPQ. Art, ddl. ■S. If the point A lies in the axis XY (Fig. 3), k't the pupil show that the same result iw obtained. q. e, d, y Google 4j4 book IX. SOLID GEOJIETRY PiiO POSITION" XXR'. Theorem 809. The area of the surface of a sphere is equal io fht prodiin of the diameter of the sphere by the circum fere.net of a gnat virde. Given a sphere generated by the revolution of the semi- circle ACE about the diaiueter AK, with the surface of the sphere denoted by S, and its radins by li. To prove S==AEX2 nli. Proof. Inscribe in the given semicircle the half of a regular polygon of an even nnmber of sides, as ABODE. Draw the apothem to each side of the semipoiygon, and denote it by a. From the vertices B, G, D draw Ji to AE. Then area AK = AF X2na.\ areaflC=F0X2 -^a. ^ r, ,^j' ■., n 1 Art. 808, area CB^0KX2 na. &reaDE==KEX2 na. ] Adding, area ABGBE^AE X 2 na. If, now, the number of sides of the polygon be indefi- nitely increased, area ABODE approaches S as a limit. Art. m. And a approaches E as a limit. (Why f) .-. AE X 2 Tta approaches AE X 2 ttK as a limit. (Why !) But area ABGDE^AE X 2 jt« always, .-. S = AEX2nR. (Why!) Q. S. B. yGoosle SPHEEICA.L AREAS 435 810. Formulas for area of surface of a sphere^ Substitutmg for AE its equal 2 E, S-4 7ti^^ Also denoting tlie diameter of the sphere by jD, E = i J). :. S=4 7l('~V, orS-TiD^. <!)' 811. Cor. 1. The surface o/g sphere is equivalent to four limes the area of a great circle of the sphere. 812. Gob. 2. The areas of the surfaces of two spheres are to each other as the squares of their radii, or of their diameters. For, if ,S and S' denote the surfaces, E and B' the radii, and D and ly the diameters of two spheres, 8 ^ 4 ^E^ ^ir- 8 ■htC- ^jy^ S' inR''^ B'2' ^'^^6" 7ii>'= D'-' 813. Property of the sphere. The following property of the sphere is useil m tlm pi oof of Art. 809: //, in tlie generat- ing arc of ami zoni. a biokcn Uiic be inscribed, whose vertices divide the arc ndo equal parts, then, as tlie number of these parts is increased indefinitely, the area generated by the broken line approaches the area of the zone (is a limit. Hence Cor. 3. The area of a zone is equal to the circumference of a great circle multiplied by the altitude of the zone. Thus the area generated by the arc BC = FO X 2 «.B. 814. Con. -t. On the same sphere, or on equal spheres, the areas of Iwo zmn-ti are to each other as the altitudes of the zones. Ex. 1. Find Ihc iii-c;i nf a sphere wliosi? difimftcr !p 10 in, Ex. 2. Fiiid tlif nni\ of a mue of alUtudir ;i io., on a sphere -whose y Google liOOK IX. SOLID PKorosiTioM XXV. Theorem 815. The area of a lime in to the urfa of (he svr~ face of ilie sjjhere as the angle of (he hine is to four riijht Given a spliere having its area denoted by B, and on the epiiere the inne ABCJ) ot £A with its area denoted by L. To prove L: S=A° -. 360°. Proof. Draw FB3, the great G whose pole is A, inter- secting the bounding area of the lime in B and D. Case I. When tlte arc BD and the circumference FB3 are commensurable. Find a common measure of BD and FBH, and let it be contained in the arc BD m times, and in the circumference FBHn times. Then arc BD : eirenmference FBII~m -. n. Through the diameter AC, and the points of division of the circumference FBH pass planes of great © . The arcs of these great ® will divide the surface of the sphere in » small equal luues, m of them being contained in the lune ABCD. :. L:8=m:n. :. L : jS=arc BD : circumference FBH. (Why ?) Or i ; «=.4° : 3G0°. Art. 257. Case II, When the arc BD and the circumference FBH are incommemurahle . Let the pupil supply the proof. q. e. ». y Google SPHERICAL AREAS 4 57 816. Fonnulafor the area of a lune in spherical degiees, or sphends. The sucfaee of a sphere contains 720 Hpheriils (Art. 803). Hence, by Art. 815, L L spherids "720 spherids that is, IJie area of a hine in 32>]ier!cal degrees is equal (0 twice the number of angular degrees in the angle of the liute. 817. Formula for area of a lune io square units of area. ■4 7it{""-^m' r„ or ;.= 818. Cor. 1. Oh ihe same Kphere, or on equal .''ijheres, two htties (ire to mch other as their anijlex. 819. Cob. 2. Two bmes with equal angles, but, on mi- equal spheres, are to each other as the squares of the radii of their spheres. '^F-A TtV'^A For L : i'.~^ : -^. or /, : L' = B- , E-. Ex. 1. Find the area in spherical degrees of a lime of 27°. Ex. 2. Find tlie number of square ioches in the urea of a lune oE 27", on a spliere whose radius is 10 in. A solid symmetrical with respect to a plane is a solid in which a line drawn from any point in ila surface X the given plane and produced its own length ends in a point ou the surface; hence Ex. 3. How many planes of symmetry has a circular cylinder f A cylinder of revolution ! Ez. 4. Haa eillier of these solids a center ot symmetry S Ex. 5. Answer the same questions for a circular cone. Ex. 6. For a cone of revolution. For a sphere. Ek. 7. For a regular square pyramid. For a regular peiitBgonul y Google 458 BOOR IS. SOLID GEOMETET pROPOsiTiox XXVI, Theorem 820. If two great circles intersect on a hemisphere, the snm of two vertical triangles thus formed is equivalent to a bme lohose angle is that angle in the triangles which is formed by the intersection of the two great circles. Given tbe liemispiiere ADBF, and on it the great circles AFB and DFC, intcracetiiig at F. To prove A AFC + A BFD ^ luue whose Z is BFV. Proof. Complete the sphere aitd produce the given arcs of the great circles to intersect at F' on the other hemisphere. Then, in the A ^FCand BF'D, araAF^RVcBF, (each Uing the siipplemenl of the arc BF). In like manner arc CF=are DF'. And arc AC—ara J)B. : A AFC~ A BF'B. Art. 79a. Add the A BFD to each of these equals; .-. A AFC + A BFD^A BF'D + A BFD. Ai, 3. Or .-. A AF(7+A£FZ)o:luue fBFJ>. Ai. 6, Q. £. I). y Google SPHEBICAL AEEAS 459 Proposition XXVII. Theoreh 821. The niiinher of spherical dr-grees, or spherids, in the area of a spherkal friaugle in equal to (he nnmber of angular degrees in the spherical excess of the Irknigle. Given the splierical A ABO whose A are denoted by ^, B, C, and whose spherical excess is denoted by E. To prove area of A ABC =i; spherids. Proof. Produce the sides AC and BC to meet AB pro- duced in the points D and F, respectively. AXBC + A Gl>B = bnie ADDC = 2 A spherids. ) > Art. 816. AABC+AACF=-hmQ£CI''A := 2 .0Brlieriils. ) A ABC + A CFP= luue of Z BOA ^ 2 C spherids. Art. >i2D- Adding, and observing that A ABC+A CDJt+AACF -I- A (7FD = hemisphere ABDFG, 2 A ABC+ hemisphere^? {A + B+ C) spherids. Or, 2 AAS(7+3COsphei-ids=2(.l + 7?+ C) spherids. Ax. s. .-. A ABC + 180 spherids^ ( A +h+ C) splierids. Ax. 5. .-. A ABC=(A + B+ C—-iiiO) spherids. Ax. 3. Ur area A ABC=E spherids. Art. Tss. y Google 4(;0 ISOOK 1\. SOLID OKOlIETilV 822. Formiila for area of a spherical triangle m square units of area. Comparing the area o£ a spheri<'ul A with the area o£ the entire spliei-e , area A : 4 %]^ = E splierids : 720 spiierids. 4 7tR= X 7J . Tin-JE :. area A = ^^ , or area A~— rr-r-- 823. The spherical excess of a spherical polygon is the sum of tlie angles of the polygon diminished by (h— 2) 180°; that is, it is the sum of the spherical excesses of tiie tri- angles into which the polygon may be divided. Proposition XXVIII. Theorem 824. The area of a spherical polygon, in spherical de- grees or spherlds, is equal to the spherical excess of the Given a spherical polygon AHCDF of n aides, with its spherical excess denoted by E. To prove area of ABGDF^E spherical degrees. Proof. Draw diagonals from A, any vertex of the poly- gon, and thus divide the polygon into n—2 spherical A. The area of each A= (sum of its i— 180) spherids. Art. 821. .■. sum of the areas of the A= [sum of /i of the ^ — («— 2) 180] spherids. As. 2. .-. area of polygon = E spherids, Art. 823. i/or the Diwi I'f A of the A— (.?i— 2) iau°=£°). y Google SPHEIUCAL VOLUMES SPHERICAL VOLUMES 825. A spherical pyramid is a por- tion of a spliere bounded by a spheri- cal polygon and the planes of the great circles forming the sides of the polygon. The base of a spherical pyramid is the spherical polygon bounding it, and the vertex of the spherical pyramid is the center of tlie sphere. Thus, in the spherical pyramid 0-ABCD, the base i ABCD and the vertex is 0. 820. A spherical wedge (or ungula) is the portion of a sphere hounded by a lune and the planes of the sides of the lune. 827. A spherical sector is the portiou of a sphere gene- rated by a sector of that semicircle whose rotation generates the given sphere. 828. The base of a spherical sector is the zone gene- rated by the revolution of the ai'c of the plane sector which generates the spherical sector. Let the pnpil draw a spherical sector in which the base is a zouc of one buse. y Google 4t)l2 T!(K)K IX, WLH> (iEOMETKY 829. A Spherical segment is a portion of a sphere ineluded between two parallel planes. The bases of a spherical segment are the sections of the sphere made by the parallel planes whlt:h bound the given segment; the altitude is the perpendicular distance between the bases. 830. A spherical segment of one base is a spherical seg- ment one of whose bounding planes is tangent to the sphere. Pkoposition' XXTX. Theorem 831. The voluuu of a -ipheif is equal to oic-ihird the product of the area of its suifiice hy its radiU"- Given a sphere having its volume denoted by F, sur- face by 5, and radius by R. To prove 7=iSXK. Proof. Let any polyhedron be eireurascribed about the sphere. Pass a plane through each edge of the polyhedron and the center of the sphere. These planes will divide the polyhedron into as many pyramids as the polyhedron has faces, each pyramid hav- ing a face of the polyhedn)H for its base, the center of the y Google SPHERICAL VOLUMES 4R^ Sphere for its vertex, siiid the ra<lius of the sphert^ for its altitude. .'. volume of each pj'ramid = i base X R. (Why ?) ,'. volume of polyhedron = 4 (surface of polyhedron) XK. If the number of faces of the polyhedron be increased indefinitely, the volume of the polyhedron approaches the volume of the sphere as a limit, and the surface of the polyhedron approaches the surface of the sphere as a limit. Hence the volume of the polyhedron and h (surface of the polyhedron) X R, are two variables always equal. Hence their limits are equal. Or "F-i«Xfi. (Why?) 832. Formulas for volume of a sphere. Substituting ^-4 7IE^ or S = 7il>- in the result of Art. 831, v= — r— ; also 7=-— -- S8S. Cor. 1. The volumes of two spheres are to each other as the cubes of their radii, or as the cubes of their diameters. 834. CoK. 2. The volume of a spherical pyramid is "/H'tl to one-third the product of its base by the radius of thr .yyhere. 835. COK. 3. Th<' roliioie of spherical ^e'ior is equal to ■.<„>--lhi,-d the prodKcl of (7.S- haxe (the bomdiiuj zone) by the i-<idnis of (he sphere. y Google 4G4 SOLID GEOJIETRY 836. Formula for the volume of a spherical sector. De- noting the altitude of the sector by R ami the volume by T, F= J ( area of zone ) X R, = J (2 7iRH) n. Art. 8U. PROrosiTiox XXX. Theorem 837. The volume of a spherical segment is equal to one- half the product of its altitude by the sum of the areas of its bases, plus the coliime of a sphere whose diameter is the attitude of tli£ segment. Given the semieirele ABCA' which generates a sphere by its rotation about the diameter AA'; BD and CF semi- chords X AA', and denoted by r and r'; and I>F denoted by-ff. To prove volume of spherical segmeut generated by BCFD, or r=h{7ir^ + 7tr'-) R-{- \ TtiP. Proof. Draw the radii OB and OC. Denote Of by h, and OD by k. Then 7=vol. 0B(7+ vol. OOP— vol. OBB. :. y=^ Ttlt-U+^h Ttr'-h—h nr-k. Arts. 830, 723. But U=h--li, A^=i^^r'^ andF^B-/--. (Whyf) y Google SPHEIilCAL VOLUMES 465 = i 7t [2 Sf {li-k)-i (B-S') *— {B"-ri (■]. Ai. a. = i Jt [2 E? (;i-f)+ H? (b-D-Ui'-fn- = iKj[3ff-(P+M+F)]. 4'— 2 Ji+F-iP. A-i. 4. ^3 B-|(,.= +, .»)-?." r-j jiH [l ( .-. r-i (rtv=+jt.-=)ir+S nH'. Q. E. B. 838. Formula for volume of a spherical segment of one base. lu a spherical segment of one base r'~o, and )'"- = (2K-ff)ir(Art. 343). Substituting for r and r' these values in the result of Art. 837, '■-"'(^-t)- 839. Advantage of measurement formulas. The student should observe carefully that, by the results obtained in Book IX, the iceasurenieut of the areas of certain curved sui-faees is reduced to the far simpler work of the measure- ment of the lengths of one or more straight lines; in like manner the measurement of certain volumes bounded by a curved surface is reduced to the simpler work of linear measurements. A similar remark applies to the results of Book VIII. Ex, 1. Find llip volume of a spheve whoso radius is T in. Ex. 3. Find tlia voluiHe of a sphere wliose diompt^r is 7 in. Ex. 3. In a spliore wIiohb radius is 8 in., find the volume of a spheriful segmeiit of one base wliosu aUitiide is 3. y Google 466 HOOK TX. fiOT.TD r.F.OMETEY EXERCISES. CROUP 72 THEOREMS CONCRRNlXr. THE SI'lfEliE Ex. i. Of circles o£ a sphere whose pLmes pass thronph a civpn. point within a sphere, tho smallest is thnt circle whoso plai.s ia perpondieAilar to the diameter through tlio given point. Ex. 2. If a point on tiie surface of a Riven sphere is equldiRtant from three points on a given small circle of the sphere, it is the pole of the small circle. Ex. 3, It two sides of a spherical triangle are quadrants, the third side measures the angie opposite that side in the triangle. Ex. 4. If a spherical triangle has one right angle, the sum of its other two augles is greater than One right angie. Ex. 6. Thfl polar triangle of a biroctangular triangle is hirectan- gular. Ex, 7. The polar triangle of a trireotangular triangle is ideatieal with the original triangle. Ek, 8. Prove that the sum of the angles of a spherical quadri- lateral is greater than 4 right angles, and less than 8 right angles. What, also, are the limits of the sum of the angles of a spherical hexagon I Of the sum o! the angles of a spherical n-gon ? Ex. 9. On the same sphere, or on equal spheres, two birectan- gular triangles are equal if tlieir oblique angles are equal. Ex. 1 1. If one of the legs of a right spherical triangle is greate than a quadrant, another side is also greater than a qiiaiJr.ant. [Sua. Of the leg which is greater than a quadrant, take the em remote from the right angle as a pole, and describe an arc] Ex. 12. II ABC and A'B'C are polar triangles, the radius OA i perpendicular to the plane OB'C. y Google EXERCISES ON THE SPHERE 4b7 Ex. 13. On tlie same Ephere, or on equal sphecee, spherical tri- angles whoae polar triangles have equal perimeters are equivalent. Ex. 14. Given OAO', OISO', and All arcs of great circles, interseoting so that Z OJC= Z O'BA ; prove that ^OAB=AO'AI!. [SuG. Show that ZOB^I= ZO'JB.] Ex. 15. Find the ratio o£ the volume of a sphere to the volume oE a circumscribed cuho, Ex. 16. Find the ratio of the surface of a sphere to the laterr.l surtaee of a ciroumscribod cylinder of revolution; a' find the ratio of their volumes. Ex. 17. If the edge of a regular tetrahedron is denoted by a, fi the ratio of the volumes of the inscribed and circumscribed ephe^j Ex. 18. Find the ratio of the two segments into which a hen sphere is divided by a plane parallel to the base o£ the UemispliBre b at the distance jB from the ba-se. EXERCISES. CROUP 73 Sl'HEKlCAL LOCI Ex. 1. Find the locus o£ a point at a given distanne n from the Bnrface of a given sphere. Ex. 2. Find the locus of a point on the surface of a sphere that is equidistant from two given points on the surface. Ex. 3. If, through a given point outside a given sphere, tangent planes to the sphere are passed, Snd the locus of the points ot tangencj. Ex. 4. If straight lines be passed through a given Ssod point in space, and through anotiier given point other straight lines be passed perpecdienlar to the first set, ilud the locus of the Xeet oi the perpendiouiuia. y Google 4t)0 BOOK IX. SOLID GEOMETRY EXERCISES. CROUP 74 rii015LT:MS COXCl':KMXfi THE STHEliB El. 1. At a given point on a sphere, eonatruet a piano tangent to the sphere. Ex. 2. Through a given point on the siirfnco of a epliere, draw an uro of s, great oirele perpendioulac to a givsn are. Ex.3. Insoribe a circle in a given spherienl triangle. Ex. 4. Construct a spherical triangle, given its polar triangle. Given the radius, r, construct a spherical surface which shall pagg through Ex. 5. Three given points. Ex. 6. Two given points and he tangent to a given plane. Ex. 7. Two given points and be tangent to a given sphere Ex. 8. One given point and be tangent to two given planes. Ex. 9. One given point and he tangeut to two given spheres. Given the radius, r, constrni^t a spherical siirfacQ which shall be tangent to Ex. 10. Three given planes, Ex. 11. Two given planes and <jne given sphere. Ex. 12. Construct a spherical surface whiah shall pass through three given points and be tangent to a given plane. Ex. 13. Through a given straight line pass a plane tangent to a given sphere. [Suo. Through the center of the sphere pass a plane J. [tiven When is the solution iinpossiiule ? Ex. 14. Through a given point on a sphere, construct an aru of a great oirele tangent to a given small circle of the sphere. [Sue, Draw a straight line from the center of the sphere to the given point, and produce it to intevseot the plane of the small oirele, etc.] yGoosle NUMERICAL EXERCISES IN SOLID GEOMETRY For methods of facilitating immerical computations, see Arts. 4i)3-6. eXEROISES. CROUP' 7S LINES AND SURFACES OP POLYHEDRON'S Find the lateral area and total area oE a riglit prism whose Ex. 1. Ease is au. equliatoral triaiigly of edge 4 in., aud whose altitude is 15 in. Es. 2, Base is a triangle of aides 17, 12, 25, aud whose altitude is 20, Ex. 3. Base is an isosceles trapezoid, the parallel bases being 10 and 15 and leg 8, and whose altitude is 24. Ex. 4. Base is a rhomhus whose diagonals are 12 and 16, and ■whose altitude is 12. Ex. 5. Base is a regular hexagon wUh side S tt., ar.i whose alti- tude is 20 ft. Ex. 6. Find the entire surface of a reetanguiar parallelepiped 8Xl2XlGiii.; ofonepX'jXrft. Ex, 7. Of a eube whose edge is 1 ft, 3 in. Ex. 8. The lateral area of a regular hexagonal prism is 120 sq. ft. and an ed^-e o£ the base is 10 tt. Find the altitude. Ex. 9. How many square feet of tin are neeessary to line a box 20X6X4in.? Ex. 10. 1( the surfac'e of a c^ube is 1 pq.vd,, find an edge in inehes E!t, 11. Find the diagonal of a cuhti whoso edge is 5 in. Ex. 12. I£ the diagonal of a cube is 12 ft., find the surface. yGoosle 470 HOLID GEOMF.TliY Ex. 13. If tho surfaop oEnreowngiilaipnrallolopipedia 208aq. in., and the odg03 are aa 2 ; y : 4, liud tUu oJjskh. In a regular square pyramid Es. 14. IE an udge o£ the base ia Id and eUut height is 17, find the ftltitude. nteral edge is 17, find nn edge Ex. 16. If i\ lateral edge is 2:i aud an edge of the Ijiae is 14, find the altitude, In a regular ti'iangular pyramid Ex. 17. If an edge of the baso is 3 and the altitude ia 10, find tho slant height. Ex. 18. Find the altitude of a regulav tetrabudron whouo edge \i 6, Find the lateral surface and the total surface o£ Eic. 19. A regular square pyramid an edga ot wlioso base is 16, and whose altitude is 15. Ex. 22. A regular square pyramid whose slant height is 24, and ihose laterai edge is 25. Ez. 23. A regular tetrahedron whose edge ia 4. Ex. 24, A regular tetrahedron whose altitude ia 9, Ex. 26. In the frustum of a regular square pyramid the edges of the bases are 6 and 18, and the altitude ia 8. Find the slant height. Hence find the lateral area. Ei, 27. In tho fru.^tum of a regular triangular pyramid the edges ot the bases are i and 6, and the altitud. ia 5, Find the slant height. Hence find the lattirul a-ttat. y Google NUMERICAL EXERCISER IN SOLID GEOMETRY 471 Ex. 28. In tlie frustum of a regular tetrahedron, if the edge of the lower base is hi, the edge of the upper base is 63, and the altitude is a, show that L = il/i{b, — b-i)' +ia'. Ex. 29. lu the fruatum of a regular square pyramid the edges o£ the bases are 2U and 60, and a lateral edge ia 101, Find the lateral EXERCISES. CROUP ?e LINES AND SURFACES OP CONES AND CYLINDERS Ex. 1. How many square feet of lateral surface has a tunnel 100 I. long and 7 ft. in diaiueteiv In a cylinder of revolution Ex. 4. Find E in terms of S and H. Es. 6. Find H in terms of E and T. Ex. 6. Find T in terms of 5 and il. Ex. 7. How many sq, yds of canvas are required to make a coni- oal tent 20 ft. in diameter and 12 ft. liigb ! El. 8. A man has 400 aq. yds. of canvas and wants to make a conical tent 20 yds. In diameter. What will be its altitude 1 Ex. 9. The altitude of a cone of revolution ia 10 ft. and the lat- eral area is 11 times the area of the base. Find the radius of the baae. In a cone of reTOlution Ex. 10. Find T In terms of S and L. Ex. 1 1. Find E in terms of T and L. Ex. 12. How many square feet c funnel the diametere of whoso ends altitude ia 7 in. ? Ex. 13. If the slant height of a fnistnm of a eoue of makes an angle of 45" with the base, show that the lateral a y Google SOLID GEOMETRY EXERCISES, CROUP ^T RPHr;R!CAT- LINES AND SL'RrACi:^ El, 2, How many squaro inelies of leather will It take to eovar a bafieball whose diameter is 3^ in.f Ex. 3. How many aq. ft. of tin are required to covpr a dome in the shape of a liemiaphere 6 yds. in diameter ? Ex. 4. What ia the radius of a sphere whose surface is filG sq. in. f Ex. 5. Find the diameter of a globe whose surface is 1 sq. yd. Ei. 7. If a hamispherical dome is to contain 100 sq. yds. of sur- face, what must its diameter be V Ex. S. Find the radius of a sphere in which the area of (he sur- face equals the number o£ linear units in the circumference of a great Find the area of a !une in which Ex, 9. The angle of the lune is 3(i°, and the radius of the sphere is 14 in. Ex. 10. Tlie angle of the lune is lfi° 'M, and the diameter of the Ex. 11. The angle of the lune is 24°, and the surface of the sphere is i sq. ft. Find the area of a spherical triangle in which Ex. 12, Tiie angles are 80^, 90", 120°, and the diameter of the sphere ia 14 ft. Es. 13. The angles are 74° 24', 83° 16', 92° 20', and the radius of the sphere is 10. Ei. 14. The angles are 85°, S5°, 135°, and the surface of tlie epliere ia 10 aq. ft. y Google NUMERICAL EXERCISES IN SOLID GEOMETEY 473 Ex. 15. If the sides of n spherical trianKle nre 100°, 110^, 120" and the radius of the sphero U 16, find the area of the polar triangle, Ex. 16. It the angles of a spherical triangle are 00", 100". 1L>0'* and its area is Bi)00, fiud tho radius of the sphere. Ex. 17. If the area of an equilateral apherioai triangle is one- third the surface of the spliere, fled an angle of the triangle. Ex. 18. In a trihedral angle the plane angles ot the dihedral angles are 80", i>0°, 100°; find the number of solid degrees in the trihedral angle, Ex. 19. Find the area of a spherical hexagon eanh of whoaa Ex. 20. If each dihedral angle of a given pentahedral angle is 120", how many solid degrees does the pentahedral angle contain 1 Ex. 21. In a sphere whose radius is 14 in,, fiud the area of a zono 3 in, high. Ex. 22. What is the area of the north temperate zono, if the earth is taken to be a sphere vrith a radius of 4,000 miles, and tlie distance between the plane ot the arctic circle and that of the tropic of Cancer is 1,800 miles t Ex, 23. If Cairo, Egypt, is in latitude 30°, show that its parallel ot latitudft bisects the surface of the northern hemisphere. Ex. 25. How much of the earth's surface will a man see who is ,000 miles above the surface, if the diameter is taken as 8,000 milesS * of a zone equals the area of a groat circle, zone in terms of the radius of the sphere, Ex. 27. If sounds from the Krakato a explosion were heard at a distance of 3,000 miles (taken as a chord) on the surface of the earth, over what fraction of the earth's surface were they heard 1 Ex. 28. The radii of two spheres are 5 and 12 in. and their cen- ters are 13 in, apart. Find the area of the circle of intersection and also of that part of the surface of each sphere not Included by the otiier ephere. y Google 474 SOLID GEOMKTKY EXERCISES. CROUE' 7a VOLUMES OF POI.YHEDKONS Find the volume of a priam Ex. 1. Whose base is an equilateral triangle ivitli Hide 5 !n., and 1j, and whose Ex. 3. Whose bnae is an isoaeelea rigbt triangle with a leg equal to 2 yds., and whose altitude is 25 ft. Ex. 4. Whose base is a regular hexagon wvitli a side of 8 ft , and whose altitude is 10 yds. Ex. 5. Whose base is a rhombus one of whose sides is 2ii, and one of whose diagonals is 14, and whose altitude is 11. Ex. 6. Whose base contains 84 bc;. yds., and whose lateral fanes are three rectangles with areas of 100, 170, 210 sq. yds., respectively. Ex. 7. How many bushels of wheat are held by a bin 30x10x6 ft., if a tii^hel is taken as li cu. ft.? Ex. 8. How many eart-loads of earth are in a collar 30 s 20 x G ft, if a uart-load is a cubic yard ? Ex. 9. If a etibioftl block of marble coats $3, what is the cost of a tube whose edge is a diagonal of the first blocl; t Ex. II. Find the edge of a cube whose volume equals the areii of its surface. Ex. 12. If the top of a cistern is a rectangle 12 s 8 £t., how deep must the cistern be to hold 10,000 gallons F Ex. 13. Find the inner edge of a peck measure which is in the shape of a cube. Ex. 14. A peck Taeasure is to be a rectangular patallelopipod with square base and altitude equal to twioe the edge of the base. Find y Google NUMEKICAL EXEHCISES IN SOLID GEOMETRY 475 Ex. IS. Find the volume of a cube whose diagonal is a. Find the rolurae of a pyramid Ex. 16. Whose base is an equilateral triangle with side 8 in,, and whose altitude is 13 in. Ex. 17. one leg 21, Whose and wh 1 base is a right 1 ose altitude ie 20, :rianele with hypotei ™. >e 29 an ,d Efc. 18. Whose base lateral edges is 5. is a square with i iide (>, and ea. ?h of who' .e Ex. 19. Whose lateral fa^es mal^e . hase is a square ngle of 45° V with siiJe 10, tith the base. and en ith of whoi ,e Ex 20 H the pvca' re b^ise of si. tain r What mid of Memphis ha ae 2:i;!yd3., how m is this worth at $1 b IS an altitude ■any cubic ya >cu. yd.? of rds 146 yd of stoi Ex. 21. A ehiirch spire 150 ft. high is hesagonal in shape and each side of the base is 10 ft. The spire has a hollow hexagonal interior, eanh side of whose base is 6 ft., and whose altitude is 45 ft. How many oubic yards of stone does the spire contain 1 I. yds. and its base is a square Ex. 23. A heap of candy in the shape of a frustum of a regular square pyramid has the edges of its bases 2ri and !] in . and its altitude 12 in. Find the number of pounds in the heap if a pound is a rectan- gular parallelepiped 4x3x2 in. in size. Ex. 24. Find the yoluma of a frustum of a regular triangular pyramid, the edges of ths bases being 2 and 8, and the slant height 12. Ex. 25. The edgea of the bases of the frustum of a regular square pyramid are 24 and G, and each lateral edge is 13 ; Snd the volume. Ex. 26. If a stick of timber is in the shape of a frustum of a regular square pyramid with the edges of its ends and 15 in., and with a length of 14 ft., End the number of feet of lumber in the stick. What is the difference between this volume and that of a stick of the sumo length having the shape of a prism with a base equal to tie area o! a midsection ot the first sUck ; y Google Ex.. 28. How many uavt-loada o£ eartli are there in n railroad cut 12 ft. deep, n-liose base is a rectangle 100 x 8 £t., and whose top is a reetangle 30 x 50 It, I Ex. 29. Find the volume ot a prismatoid whoso baae is an equi- lateral triangle with aide 12 ft., nnd whose top is a Una 12 ft. long liarallel to one Bide of the base, and whose altitude is 15 ft. Ex, 30. If the haae of a prismatoid is a rectangle with dimensions a and b, the top is a line c parallel to the aide !i of the liase, and the altitude ia 'i, find the vulume. EXERCISES. CROUP ta VOLTTJIES OF COXES AND CYLlNDii;RS Ex. 1. How many barrels of oil are contained in a cjlicdrical tank 20 ft. long and G ft. in diameter, if a barrel contains 4 cu. ft. t Ex. 2. How many ou. yds. of earth must be removed in making a tunnel 450 ft. long, it a eross-seotion of the tunnel is a aeiEiicirole of 15 ft. radius f Ex. 3. A cylindrical glass 3 in. in diameter holds half a pint. Find its height in inches. Ex. 4. If a cubic foot of lameter, how long will the w altitude equals the Ex. 6. Showthat the YOlumesof two cylinders, having the lUtitude f each equal to the radius of the other, are to each other as E : S'. Ex. 8. A conical heap of potatoes is 44 ft. in circumfeienee and 6 ft. high. How many bushels does it contain, if a bushel is lieu. It.t e- glass hold, it y Google NTTMEEICAL EXERCISES IN SOLID GEOMETItY 477 Ex. 10. Find the ratio of tho volumea o{ the two eonts ins^iribed ribed about, a regular tetrahedron. X. 11. IE an equilateral eono eontaius 1 quart, find its diraen- b of R aud L ; alao Ejc. 13. Find tiie volume of a frustum of a cone of revolution, whose radii are 14 and T ft., and whose altitude is 3 yds. Ex. 14. What is the cost, at 50 cts. a oh, ft., of a piece of marble in the shape of a fruBtum of ft cone of revolution, whose radii are 6 and 9 ft., and whose slant height is 5 ft.? ExenciSEs. crouf so SPHERICAL VOLUJLES Ex. 1. Find the volume of a sphere whose radius is 1 ft. 9 in. Ex. 2. I! the earth is a sphere 7,9L'0 miles In diameter, Bnd iti Ex. 3. Find the diameter of a sphere whose volume is 1 cu. ft. Ex. 4. What is the volume of a sphere whose surface isClG sq. in.' Ex. 5. Find the radius of a sphere equivalent to the sum of twi pheres, whose radii are 2 and 4 in. Ex. 7, Find the volume of a sphere oircumseribed whose edge is fi. Ex. 8. Find the voluiue of a spheiifial shell whc outer diameters are 14 and 21 iu. Ex. 9. Find the volume o( u spberii.'al shell whoso ii surfaces :izo 20 tt wnd i: tt. y Google 4 1 8 SOLID GKOJIETllY Ifind tiio voluBid of Es. 10. A splierieal wedge whoso auglo is 3!", the radius of the sphere being 10 in. Ex. 11. A spherical Keetor whoso base is a. zone 2 in. high, the rartius of the sphere being 10 in. a whose rsdii are 4 and Ex. 13. A wash-basin in the shape r>£ a aogment, of a sphere is 6 in. deep and 2i in. in djonieter. How many quarts o£ water will the basin hold t Ex. 14. A plana parallel to the base of a hemisphere and bisect- ing thi! altitude diifides its volume in what ratio f Ex. 15. A aplieri<^al Kogmeut 4 in. high «oiitains 200 eu. in. ; End the r.idiu.'i of the sphere. Ex.16. If a heavy sphere whoso dia ponieal wice-Glass full of water, whose i 6 in., find how much water will run over EXERCISES. GROUP 81 EQUIVALENT SOLIDS Ex. i. IE a cubical block of putty, eaffh edge ot whi ch is 8 incheg, be molded into a cylinder oE revolution whose radius h 1 3 inches, find the altitude ot the eyiicder. Ex. 2. Find the radius of a sphere equivalent to a cube whose edge ie 10 in. Ex. S. Find the radius of a sphere equivalent to a i ;one of revolu- tion, whose radius ia 3 in. and altitude G in. Ex. 4. Find the edge of a cube equivalent to a frustum of a eons f revolution, whose radii are 4 and 9 ft. and altitude 2 yds. Ex. 5. Find the altitude of a rectangular parallelopiped, whose ase is 3x3 in. and whose volume Is eqaivaleot to a spheru of tadiua y Google NUMERICAL EXERCISES IN SOLID GEOMETlil' 479 Ex. 6. Find the biiRe of a square reetangular pai'allelopiped, Vfliose altitude ia 8 in. aud whose volume equals the volume o£ a cone of revolution with a radiuB of 6 and au aititudo of 12 m. Ex. 7. rind the radius of a cone of revolution, whose altitude is 15 and whose volume is equal to that of a eyliniier of revolutiou ivith radius G and altitude 20. Ex. 8. Find the altitude of a eone of revolution, whoso radius is 15 and whose voiume equals the volume of n cone of revolulion with radius 9 and altitude 24. Ex, 9. On a sphere whose diameter is 14 the altitude of a zone of one base is 2. Find the altitude of a pylinder of revolution, whose base equals the base of the zone and whose lateral surface equals the surface of the zone. EXERCISES. CROUF* SIMILAE SOLIDS Ex. 1. If on two similar solids L, L' and !, I' are ORous lines; -4, A' and a, «' pairs of homologous T, v' pairs of homologous volumes, pairs of homol- veas, V, V and ft :L' = 1 :I' = v'-^ ■.i/A' = ^'v.f'e'. show that - A : A' = L^ : L" = a : n'= r* : FL Ex, 2. If the edge of a cube is :o in., &nd the having 5 times the surface. edge of a cube Ex. 3, If the radius of a sphere is 1(1 in., find the radiu Bphere having 5 times the surface. Ex. 5. In the last three exercises, find the required di 10 volume ia to be 3 times the volume of the original solid. those of auotliui- trunk. How y Google SOTJD GEOMliTUy Ex. 7. How fill from Iho vertex is i Ihe croBK-section which bisects tho volume of a cone Ot I'BVOlutiOI 1 f Wliich biseeta tiie lateral Eurfaee ? Ex. 8. If the nltiliide oC a pyramid is bisected Tjy a plane parallel to the 1)sse, how does the area of tho cross-seetion ootupare with the area of the liase ? How does the volume cut olf compare with the volume of the entire pyrnmid ? Ex. 9. Plnnes parallel to Iho base of a coae divide tho altitude into three equal pai'ts ; compare the lateral surfaces cut off. Also the volumes'. Ex. 10. A sphere 10 i(i. in diameter ia divided into three equiva- lent parts by concentric spherical surfaces. Find the diameters of toese surfaces. Ex. 11. It the strength of a muscle is as the area of its cross- section, and Goliath of Gath was three times as large ia each linear dimension as Tom Thumb, how much gre;iter was his strength ! His weight f How, then, does the activity of the one man compare with that of the other ? Ex. 12. If the rate at which heat radiates from a body is in pro- portion to the amount of surface, and the planet Jupiter has a diame- ter 11 times that of the earth, how many times longer will Jupiter be in cooling off t [SuG. How many times greater is the volume, and therefore the original amount of heat in Jupiter ? How many times greater is its surface f What will be the combined effect of these factors T] CROUP 83 MISCELLANEOUS NUMERICAL EXERCISES IN SOLID GEOMETRY Find S, T and V of Ex. 1. A right triangular prism whose altitude is 1 ft., and t! Bides of whose base are 26, 28, 30 in. base ia 1ft. 2ii Bx. 3. A frustum of a square pyramid Iho areas of whose has are 1 sq. It, and 3G sq. in,, and whose altitude ia !) ia. y Google NUMERICAL EXEECISES IN SOLID GEOMETRY 48X Ex. 4. A pyramid whose slant height is 10 in., and whose base I an equilateral triangle whose side is 8 in. Ex. 6. A frustum o( a cooe of revolutioo whose radii 11 in. and slant height 13 in. Ex. 8. Find the volume of a sphere inscribed in a cube whose edge is 6; also find the area of a triangle on that sphere whose anglea are 80°, 90", 150°. > of the spherical pyramid whose base is Ex. 10. Find the angle of a lime on the same sphere, equivalent to that triangle. Ex. II. On a cube whose edge is 4, planes through the midpoints of the edges eut oE the corners. Find the volume of the solid re- maining. Ex. 12. How is F ehnnged if n of a cono of revolution is doubled and R remains unchanged t If R is doubled and W remains unchangedt If both H and J! are doubled ? Ex. 13. In an equilateral cone, find S and V in terms of E. Ex. 14. A piece of lead 30 x H :t 3 in. will make how many spher- ical bullets, each S in. in diameter ! Ex, 16. How many bricks are necessary to make a chimney in the shape of a frustum of a eone, whose altitude is DO ft., whose outer diameters are 3 and 8 ft,, and whose inner diameters are 2 and 4 ft., eonnting 12 bricks to the eubio ft. ? a of a zone is 300 and its altitude 6, find t Ex. 18. If every edge of a square p7iamidisit Qtid^in termsof T. y Google 453 SOLID GKOMliri'llI: Ex, 19, Arpgiikr ^qHarc pyramid hns a for its allituJe and also t ,,. encli side of the liiu-e. Find llie area of a section made by a plane parallel to the base mid bifieGting Ilia altitude. Find also tlie volumea of tha two parts into wliieli the pyramid ia divided. Ex. 20. It the earth is a sphere of 8,000 milos diameter and its alroosphere extends 50 miles from the eavlb, find the volume of the Ex. 21. On a spherp, find the ratio of llie area or an e.iuilateral ipberipal Iriiinglp. eaeh of wliose angles ia 35', to liie area of a lune Ex. 22. A square right prism has an altitude Cm and an edge o£ the base 2a. Find the volume of the largest cylinder, sphere, pyramid ami cone which can be ent from it. Ex. 23. Obtain a formula for the area of that part of a'sphere illujuinateit by a jioint o£ light at a distance a from the sphere whose radins is H. Ex. 24. On a sphere whose radius is 6 in., find an angle of an equilateral tnaBjtle whose area is 12 sq. in. Ex. 25. Find the volume of a priemiitoid, whose altitude is 2-1 and whose bases are equilateral triangles, each side 10, so placed that tlie raid-seotion of the prismatoid is a regular hexagon. Ex. 26. Ou a sphere tihose radius is 16, the bases of a zone are equal and are together equal to the area of the zone. Find the alti- tude of the zone. e of a square pyramid, the edge of whose teral edges is inclined G0° to the base. Ex. 28. Aq irregular piece of ore, if placed in a cylinder partly filled with water, causes the water to rise 6 in. If the radius of the cylinder is 8 in., what is the volume of the ore ? Ex. 29. Find the volume of a truneated right triangular prism, if the edges of the base are 8, 9, 11, and the lateral edges are 12, 13, 14. Ex. 30. In a sphere whose radius is 5, a aectioii is taken si tho distance 3 from the center. On this section as a base a oone iu formoti whose lateral elements are tangent to the sphere. Find tho lateral surface and volume of the coua. y Google NUMERICAL EXEECIRES IS SOLID GEOatETRY 483 Ex.31. The volume of a sphere 13 l,437iau. in. Find the surface. Es. 32. A square whose side is 6 is revolved about a diagonal as a axis ; find the surface and volume generated. Ex. 33. Find the edge ot a cubical cistern that will hold 10 tons f water, if 1 cu. ft. of water weighs 62.28 lbs, Ex. 34. A water trough has equilateral triangles, eaeh side 3 ft., )r ends, and is IS ft. loiip. How many buckets of water will it hold, ! a bucket ta a cylinder 1 ft. in diameter and li ft. high J Ex. 35. Tlie lateral ar. ea of a (.-yiinder of r evolution is 440 sq. d the volume is 1,5411 cu .. in. Find the ra.lli IS and altitude. Ex. 36. The anfrles of a spherical quadrilateral are 80", 100", 1-0", 120°. Find tile angle ot an equivalent equilateral triangle. Ex.37. A cone and a cylinder have equal lateral suifacep, and their axis sections are equilateral. Find the ratio of their volumes. Ex.38. A water-pipe I in. 1 How many quarts of water mu from under the ground oemes i how long must the watei' run f Ex. 39. A cube immersed causes the water to rise 4 in. what is an edge ot the cube 1 Ex. 40. An auger hole whos Ex. 42. The volumes ot two similar cylinders of revolution are as 8 : lliS; find the ratio of their radii. If the r.tdius of the smaller if 10 in., what is the radius of the larger? Ex. 43. An iron shell is 3 in. thick and the di^imeter of Us oute: surface is 28 in. Find its volume. Ex. 44. The legs of an isosceles spherical triangle each make fir ".ngle of Ta° «ith the base. The legs [>i'odueed form a iiine whosi aiea is four times the area of the triangle. Find the angle of the lune y Google 484 SOLID CEO^iirrsY CROUP 81. EXERCISES INVOLVING THE MISTRIC S^YSTEM rind s. r, r o( Ex. 1. A riKht priam tho edRes of nhoso bas^o are li m., 70 dm., 900 cm., and wbose altitude is 90 dm. Ex. 2. A regular square pyramid sn fidge of whose base is 30 lim., and whose altitude is 1.7 m, Ex. 3. A sphere whose radius is 0.02 m. Ex. 4. A frustum of a oone of reTolution whose radii are 10 dm acd 6 dm., and whose slant height ia 50 cm. Es. 5. A cube whose diagonal is 12 cm. Ex. 6. A cylinder of revolution whose radius equals 2 dm., auii ■whose altitude equals the diameter of the base. Ex. 7. Find the area of a sphericnl trianglp on a sphere who^f radius is 0.02 m., if its angles are 110°, 120°, 130°. Ex. 8. Find the number of square meters iu the surface of [ sphere, a great circle of which is 50 dm. long. Ex. 9. How many liters will a cylindrical vessel hold that i 10 dm. in diameter and 0.iJ5 m. high ? How many liquid quarts ? a cylinder whose diameter is half th. Ex. 1 1. Find the surface of a sphere whose TOlume is 1 cu. m. y Google APPET^DIX I. MODERN GEOMETRIC CONCEPTS 840. Modern Geometry, In recent times many new georaetfie ideas have been invented, and some of them developed into important new branches of geometry. Thus, the idea of symmetry (see Art. 484, etc.) is a modern geometric concept. A few other of these modero concepts and methods will he briefly mentioned, but their thorough consideration lies beyond the seope of this hook. 841. Projective Geometry. The idea of projections (see Art. 345) has been developed in eomparatiyely recent times into an important branch of mathematics with many praetieal applications, as in engineering, architec- ture, constructicm of maps, etc. 842. Principle of Continuity, By this principle two or more theorems are made special cases of a single more general theorem. An important aid in obtaining continuity among geometric principles is the application of the con- cept of negative quantity to geometric magnitudes. Thus, a negative line is a line opposite in direction to a given line taken as positive. For example, if OA is +, OB is - (485) yGoosle 48G Similarly, a negative angle is an angle formed by rotating a line in a plane in a direction opposite from a direction of rotation taken as positive. Thus, if the line OA rotating from the position OA forms the positive angle AOB, the same line rotating in the opposite direction forms the negative angle AOB'. Similarly, positive and negative ares are formed. In like manner, if P and P' are on opposite sides of the line AB and the area PAB is taken as positive, the area P'AB will be negative. As an ilhistratiou of the law of continnity, we may take the Uieoreni that the sum of the triangles formed by drawing lines from a point to the vertices of a polygon equals the area of the polygon, " " Applying this to the qnadrilateral ABCD, if the point Pfalla within the quadrilateral, ^PAB+AFBC + APCI) + i^rAD=ABCr> (Ax. 6). Also, if the point falls without the quadrilateral at J", AP'AB + AP'BG+ AP'C7>+ A P'AD == ABCD, since APAD is a negative area, and hence is to be subtracted from the sum of the other three triangles. 843. The Principle of Reciprocity, or Duality, is a principle of relation between two theorems by which each theorem is convertible into the other by causing the woi-ds for the same two gooniotrie objects in each tlicorem to ex- cliaTige plaecK. y Google MODERN GEOMETRIC CONCEPTS 487 Thus, of t.liL'orems VI and VII, Book I, either may be convei'ted into the otiier by replacing the word "sides" by "aogles," and "angles" by "sides." Henee these are termed reciprocal theorems. The following are other instances of reciproeiil geometric properties : 1. Two points 2. Three p<ri«U straight line deleriN determine a <e a plane. 1. Tw 2. Tin lines lie t ermine a point. ee plaim 7inl through the ight line ileierniine a 3. A straight U AeUnnine a piune. " <""' « I'O'"' 3. Js deter iiiiiw fraighl line wul a plane- a i'oiiit. The reciprocal of a theorem is not necessarily true. Thns, two parallel straight lines determine a plane, but two parallel planes do not determine a line. However, by the use of the principle of reciprocity, geometrical properties, not otherwise obvious, are fre- quently suggested. 844. Principle of Homology. Just as thii law of reci- procity indicates relations between one set of geometric concepts (as lines) and another set of geometric concepts (as points), so the law of homology indicates relations between a set of geometric concepts and a set of concepts outside of geometry: as a set of algebraic concepts, for instance. Thus, if a and l are nnmbers, by algebra {a + h) (a — i) Also, if a and b are segments of a line, the rectangle («+6)X (i( — h) is equivalent to the difference between the squares ir and V'. By means of this principle, truths which would be over- looked or difficult to prove in one departmeut of thonght y Google 488 (lEO^liriKY. Al'I'ENVlX are made obvious by observing tlie corresponding truth in another department of thought. Thus, if a and h ai-e line segments, tlie theorem (a + h)" + {a—b}- — 2{(t--i-b-} is not immediately obvious in geo- metry, but becomes so bs observing the Hke relation between the algebraic uumbers a and i>. 845. Non-Euclidean Geometry. Hyperspace. By vary- ing the properties of space, as these are ordiuarily stated, different kinds of space may be conceived of, each having its own geometric laws and properties, Thus, space, as we ordinarily conceive it, has three dimensions, but it is possible to conceive of space as having four or more dimensions. To mention a single property of four dimen- sional space, in sneh a space it would be possible, by simple pressure, to turn a sphere, as an orange, inside out without breaking its surface. As an aid toward eonoeiving how tliia ia poaslblo, consider a plane in wiiicil one eii'cle lies inside another. No matter how these eirelea are moved about in the plane, it is impossiljle to shift the inner eirale KO aa to piaee it outside the other witliout l)rea!iing the oiroumference III the outer circle. But, if we are allowed to use the third dimension of space, it ia a aimple matter to lift the inner circle up out of the plane and set it down outside the larger oirale.' Suuilarly it, in space o£ three dimensions, we have one spherical aliell inside a larger shell, it is impossible to place the smaller shell ontsida the larger without breaking the larger. But if the nse of a fourth dimension be allowed, — that is. the use of another dimension of freedom of motion, — it ls possible to place Che inner shell outside the larger without breaking the latter. 846. Curved Spaces. By varying the geometric, axioms of space (see Art. 47), different kinds of space may be conceived of. Thus, we may conceive of space such that through a given point one line may be drawn parallel to a given ime (that is ordmary, or Euclideau space}; or such y Google MODF.KS GEOMETillC COKCEPTS 489 that through a given point no line can be drawn parallel to a given liue (spherical space); or such that through a given point more than one line can be drawn parallel to a given liue (pseudo- spherical space). These different kinds of space differ in many of their properties. For example, in the first of them the sum of the angles of a triangle equals two right angles; in the second, it is greater; in the third, it is less. These different kinds of space, however, have many properties in common. Thus, in all of them every point in the perpendieaiar hiseetor of a liue is equidistant from the extremities of the liue. eXERCISKS. CROUP SE Ex. 1. Show by the use of zero and negative ares that the princi- ples of Arts. 257, 263, 258, 264, 203, are particular cases of the genei'al theorem that the angle inoluded between two lines which cut or tooth a cirule is measured hj oue-lialf the sum of the intercepted area. Ex. 2, Show that the principles of Arts. 354 and 3o8 are particular eases of the theorem that, if two lines are drawn from or through a po'nt t m t a eireumfa ranee, the product o£ the segments o£ one line equ h p duct of the segments of the other iino. Ex 3 bhow by the use o£ negative angles tha h em XXXVIII, Book I, is true for a quad a e al o£ the form AISCD. IBGD is a ne a e an le; the angle at the vertex i» is the fl sa leADC] Ex 4 What is the reciprocal of tiie state- Ex. 5, What is perpendicular to i dicular to eiich oti y Google IT. ITIRTORY OP GKO:\IETRY 847. Origia of Geometry as a Science. The beginumgR of geometty as a science are found in Kgypt, diitiug back fit least tliree thonsaud years before Christ. Herodotus says that geometry, as known iu Egypt, grew out of the need of remeasuring pieces of laud parts of which had been washed away by the Xiie floods, in order to make an e^ni- tabie readjustment of the taxes on the same. The substance of the Egyptian geometry is found in an old papyrus roll, now iu the British museum. This roll is, in eSect, a mathematical treatise written by a scribe named Ahmes at least 1700 B.C., and is, the writer states, a copy of a more ancient work, dating, say, 3000 B. C. 848. Epochs In the Development of Geometry. Prom Egypt a knowledge of geometry was transferred to Greece, whence it spread to other countries. Hence we have the following principal epochs in the development of geometry: 1. Egyptian : 3000 B. C— 1500 D. C. 2. Greek : COO B. C— 100 B, C. 3. Hindoo : 500 A. D— 1100 A. D. 4. Arab : 800 A. D.— 1200 A, D. 5. European : 1200 A. D, In the year 1120 A. D.. Athelard, an English monk, visited Cordova, m Spam, in the disguise of a Mohamme- dan student, and proeui-cd a copy of Euclid m the Arabic language. This book he brought back to central Europe, where it was translated into Latin and became the basis of all geometric study in Europe till the year lriJ3, when, (WO) yGoosle niKTORY OF GEOMETRY 4i'l owing to the capture of Ooiiatautiiiople by the Turks, copies of the worlis of the Greek mathematicians in the original Greeli were scattered througli Europe. HISTORY OF GEOMETRICAL METHODS 849. Rlietorical Methods. By rhetorical methods in the presentation of geometric truths, is meant the use of definitions, axioms, theorems, geoiiieti'ic figures, the rep- resentation of geometi'ic magnitudes hy tlie use of letters, tlie arrangement of material in Books, etc. The Egyptians iiad none of these, their geometric knowledge being re- corded onlj' in the shape of the solutions of certain numeri- cal examples, from which the rules used must be inferred. Thales (Greece COO B.C.) fii'st made an enunciation of an abstract pro]»ei'ty of a gcometi'ie figure. He had a rude idea of the geometric theorem. Pythagoras (Italy 535, B.C) introduced formal defini- tions into geometry, though some of those used by him were not very accurate. Fof instance, his definition of a point is "unity having position." Pythagoras also arranged the leading propositions known to him in something like logical order. Hippocrates (Athens, 420 B. C.) was the tirst systemati- cally to denote a point by a capital letter, and a segment of a line by two capital tetters, as the line AB, as is done at present. He also wrote the first text-book on geometry. Plato (Athens, 380 B.C.) made definitions, axioms and postulates the beginning and basis of geometry. To Euclid (Alexandria, 280 B. C.) is due the division of geometry into Books, the formal enunciation of tlico- rems, the particular enuucialion, the f(>nual constru(;rion, y Google 492 GEOMETRY. Arl'ENDrX proof, and cuuclusioii, Ui presenting a proposition. He also introduced tlie use of the corollary and scholium. Using these methods of presenting geometric truths, Euclid ■wrote a text-book of geometry iu thirteen books, which ■was the standard text-book on this subject for nearly two thousand years. The use of the symbols A, CO , \\ , etc., in geometric proofs originated in the United States iu recent years. 850. Logical Methods. The Egyptians used no formal methods of proof. They probably obtained their few crude geometric processes as the result of experiment. The Hindoos also used no formal proof. One of their writers on geometry merely states a theorem, draws a figure, and says "Behold I " The use of logical methods of geometric proof is due to tiie Greeks. The early Greek geometricians used experi- viental methods at times, in order to obtain geometric truths. For instance, they determined that the angles at the base of an isosceles triangle are equal, by folding half of the triangle over on the altitude as an axis and observ- ing that the angles mentioned coincided as a fact, but without showing that they must coincide. Psrthagoras (i325 B. C.) was the first to establish geo- metric truths by systematic deduetion, but his methods were sometimes faulty. For instance, he believed that the converse of a proposition is necessarily true. Hippocrates (420 B. C.) used correct and rigorous de- duciion iu geometric proofs. He also introduced specific varieties of such deduction, such as the method of reduc- ing one proposition to another (Art, 296) , and the reductio ad ahsurdtim. y Google HISTORY OF GSOMETKY 493 The methoda o£ deduction used by the Greeks, ho^vaver, were de- fective in their lack of generality. For instance, it was often thought necessary to have a separate proof of a theorem for each different kind of figure to which the theorem applied. Thus, the theorem that the sura of the an- y-~~~~7\ gles of a triangle equals two right angles ,-' 'v, /' '\ (1) for the equilateral triangle Tiv use '•■ / .. . . v of the regular hexagon ; (2) for the right triangle by the use of a rectangle; (3) tor a Boalene triangle by dividing the scalene triangle into two right triangles. The Greeks appeared to fear that a general proof might be vitiated if it were applied to a figure in any way special or peculiar. In other words, they had no conception of the principle o( eontinuity (Art. 843). Plato (380 B. C,} introduced the metliod oE proof by ait<'!ysi'^. that is, by taliing a proposition as true and work- ing from it back to known truths (see Art. 196) . To Eudozus (3G0B. C.) is virtually due proof by the method of limits; though his inetliod, known as the method of exhaustions, is crude and cumbersome. ApoUonius (Alexandria, 225 B. C.) used projections, transversals, etc., which, in modern times, have developed into the subject of projective geometry. 851. Mechanical Methods, Tlie Greeks, in demonstra- ting a geometrical theorem, usually drew the figure em- ployed in a bed of sand. This method had certain advan- tages, but was not adapted to the use of a large audience. At the time when geometry was being developed in Greece, the interest in the subject was very general. There was scarcely a town but had its lectures on the subject. The news of the discovery ol a y Google 4',)i GEOJIETKY. .VPI'EyDIX new theortm spreati from town to town, and the theorem nas redenjon- strated in the sanii of each raarket place. The (ireek treatises, however, were written on velhim or papyrus by tiie use of the reed, or calamus, and ink. In Roman times, and in the middle ages, geometrienl figures were drawn in wax smeared on wooden boards, called tablets. They were drawn by the nse of the stylus, a metal stieii, pointed at one end for making marks, and broad at the other for erasing marks. Tliese was tablets were stiil in nse in Shakespeare's time (see Hamlet Act 1, Sc. 5, 1, 107). The blackboard and crayon are modern inventions, their nse having developed within the last one hundred years. The Greeks invented many kinds of drawing instruments for tracing various curves. It was due to the inflneuee of Plato (1180 B. C.) that, in constructing geometric figures, the use of oniy the rnier and compasses is permitted. HISTORY OF GEOMETRIC TRUTHS. PLANE GEOMETRY, 852. Rectilinear Figures. The Egyp- tians measured the area of any fonr- sided field by multiplying half the sura of one pair of opposite sides by half the sum of the other pair; which was equivalent to using the lormuia, area — — ^X -n ~' This, of ooursB, gives a eorreet result loc the rectangle and square, but gives too great a result for otlier quadrilaterals, as the trapeaoid, etc. Hence Joseph, of the Book of Genesis, in buying the fields of the Egyptians for Pharoah in time of famine by the use of this (ormuia, ill many onses paid for a larger field than lie obtained* The Egj-ptians had a special fondness for geometrical constructions, probably growing out of their work as temple y Google HISTORY OF GEOMETRY 495 biiilders. A ekss of workers existed among them called "rope-stretchers," whose business was the marking out of the foundations of buildings. These meu knew how to bisect an angle and also to coustruct a right angle. The latter was probably done bj' a method essentially the same as forming a right triangle whose sides are three, four and five units of length. Ahmes, in his treatise, has various constructions of the isosceles trapezoid from different data, Thales (600 B, C.) enunciated the following theorems: If two straight lines intersect, the opposite or vertical angles are equal; The angles &i the base of an isosceles triangle are equal; Two triangles are equal if two sides and the included angle of one are equal to two sides and the included angle of the other; The sum of the angles of a triangle equals two right angles; Two mutuallj" equiangular triangles are similar. Thales used the last of these tiieorems to measure the height of the great pyramid by measuring the length of the shadow cast by the pyramid and a!so measuring the length of the shadow of a post of known height at the same time and making a proportion between these quantities. Pythagoras (525 B. C.) and his followers discovered correct formulas for the areas of the principal rectilinear figures, and also discovered the theorems that the areas of similar polygons are as the squares of their homologous sides, and that the square on the hypotenuse of a right triangle equals the sum of the squares on the other two sides. The latter is called the Pythagorean theorem. They also discovered how to constnict a square equivalent to a given parallelogram, and to divide a given line in mean and extreme ratio, y Google 496 GEOMETRY. Ari'ENDIX To EudoxuB (380 8. C.) we owe the general theory of proportion in geonietr.v, and the treatment of incommen- surable quantities by the method of Eshaustions. By the use of these he obtained such theorems as that the areas of two circles are to each other as the squares of their radii, or of their diameters. In the writings of Hero (Alexandria, 125 B. C.) we first find the formula for the area of a triangle in terms of its sides, K=Vs(s — a) (s — b) (s — e). Hero also was the first to place land-snrveyiug on a scientific basis. It is a curious fact tliat Haro at the same time gives an ineorreet formula for the area of a triangle, ¥iz., K=ia{i+':), tliis formula being apparent!}' derived from Egyptian sources. Xenodorus (150 B. C.) investigated isoperemetricai figures. The Romans, though they excelled in engineering, ap- parently did not appreciate the value of the Greek geom- etry. Even after they became acquainted with it, they continued to use antiquated and inaccurate formulas for areas, some of them of obscure origin. Thus, they used the Egyptian formula for the area of a quadrilateral, K=—^ X— ^. They determined the area of an equilat- eral triangle whose side is a, by different formulas, all incorrect, as K=~~- , K=i{a"+a), and ZL = ia^. 853. The Circle. Thales enunciated the theorem that every diameter bisects a circle, and proved the theorem that an angle inscribed in a semicircle is a right angle. To Hippocrates (420 B. C.) is due the discovery of nearly all the other principal properties of the circle givaa in this book. y Google HISTORi" OP GEOMETRY 497 The Egyptians i-pgarded the area of the circle an eL|iii\ni- lent to ,ff of the diameter squared, which would make T-3.iC04. The Jews and Babylonians treated 7t as equal to 3. Archimedes, by the use of inscribed and circumscvihed regular polygons, showed that the true value of 7t lies between Sf and 3ki; that is, between 3.14285 and 3.1408. The Hindoo writers assign various values to 7t, as 3, 3b, l/lO, and Aryabhatta (530 A. D.) gives the correct ap- proximation, 3.1416. The Hindoos iised the foi'mnla 1/2 1/4 A& '^^^ ■^''*- *^^^ ^^ computing the numeri- eal value of ti. Within recent times, the value of n has been computed to 707 decimal places. The use of the symhoi n for the ratio of the circum- ference of a circle to the diameter was established in mathematiea by Euler (Germany, 1750). HISTORY OF GEOMETRIC TEDTHS, SOLID GEOMETRY 854. Polyhedrons. The Egyptians computed the vol- umes of solid figures from the linear dimensions of such figures. Thus, Ahmes computes the contents of an Egyp- tian bani by methods which are equivalent to the use is not known, it is not possible to say whetlier this formula is correct or not. Pjrthagoras discovered, or knew, all the regular poly- hedrons except the dodecahedron. These polj-hedrons were supposed to have various mugica! or mystical properties. Heuce the study of them was made very iiromiaeut, y Google 49S GKOMKTRV. APl'ESinX Hippasus {470 B.C.) discovered the dodecaliodrnn. but, he was drowned by the other Pythagoreans for boasting of the discovery. Eudoxus (380 E. C.) showed that the volume of a pyra- mid is equivalent to one-third the product of its base by its altitude. E. F. August (Germany, 1849) introduced the prisma- toid formula into geometry and showed its importance. 855. The Three Round Bodies. Eudoxus showed that the volume of a cone is equivalent to one-third the area of its base by its altitude. Archimedes discovered the fonuulas for the surface and volume of the sphere. Menelaus (100 A. J).) treated of tlie properties of spherical triangles. Gerard (Holland, 1620) invented polar triangles and found the formulas for the area of a spherical triangle and of a spherical polygon. 856. Noa-Euclideaa Geometry. The idea that a space might exist having different properties from those which •we regard as belonging to the space in which we live, has occurred to different thiuiers at different times, but Lobatchewsky (Russia, 1793-1856) was the first to make systematic use of this principle. He found that if, instead of taking Geom, Ax. 2 as true, we suppose that through a given point in a plane several straight lines may be drawn parallel to a given line, the result is not a series of absur- dities or a general reductio ad absurdum; but, on the con- trary, a consistent series of theorems is obtained giving the properties of u space. y Google ITT. REVIEW EXERCISES EXERCISES. CROUP se REVIEW EXERCISES IN PLAXE GEOMETEY Ex. 1. It the bisectors of two adjacent angles are parpendioulaj to each otber, the angles ai'e supplementary. Ex. 2. If a diiigonal of a quadrilateral bisects two o£ its angles, the diagonal bisects the quadrilateral. Ex. 3. Through a given point draw a secant at a given distance ■om the center of a given circle. Ex. 4. The bisector of one angle of a triangle and of an exterior ngle at another vertex form an angle which is equal to one-half thtj lird angle of the triangle. Ex. 5. The side ol a square is 18 in. Find the ulrcumference of the inscribed and eiroumseribed circles. Ex. 6. The quftdrilateral jlDiJC iK inscriHeii in a circle. The diag- o.iiils Af! and iJCintersect in the point F. Arc ^J> = 112°, Rrc AC = lAFC=^ li". Find all the other angles of the figure. Ex. 7. Find the locus of the center of a circle which touehee two given equal circles. Ex. 8. Find the area of a triangle whose sides are 1 m., 17 dm.,, 210 em. Ex. 9. The line joining Ihe midpoints of two radii is porpendiculaf to the line bisecting their angle. Ex. 10. If a quadrilateral be inscribed in a circle and its diag- onals drawn, bow many pairs of Bimilar triangles are focnied t Ex. 11. Prove that the sum of the exterior angles of a polygon (Art. 172) equals four right angles, by the use of a ligure formed by drawing lines from a point within a polygon to Uie vortices of the polygon. (430) yGoosle GE05IETEY. Ex. 12. In a circlo whose radius is 12 em., find the langth of the lugent drawn from a point at a distance 240 mm. from the center, Ex. 13. If two sides of a regular pentagon he prodiieed, End the Ex. 14. In the parallelogrEm ABCD, points are taken on the diasoaals such that AP=BQ=CE==DS. Show that PQBS is a parailelogram. Ex. IB. A chord in. long is at the distance 4 in. from the eentcv of a circle. Find the distance from the center of a aiiord 8 in. long. Ex. 16. If B is a point in the eireumferenee o£ a eicele whosa center is 0, PA a tangent at any point P, mooting OB producBd at .-), and PD perpendicular to OH, then. PB bisects the angle AFD. Ex. 17. Construet a parallelogram, given a side, an angle, and a diagonal. Ex. 18. Find in inches the sides of an isosceles right triangle whose area is 1 sq. yd. Ex. 20. 3f two lines intersect so that the product of the segments of one line equals the product of the segments of the other, a cir- cumference may he passed through the extremities of the two lines. Ex. 21. Find the locus of the TerticBB of all triangles on a given base and having a given area. Ex. 22. On the figure p. 206, prove that^'*+4p = ZB^H-f(^^- reetangle is 108 and the base is throe times Ex. 24. If, on the sides AG and BC o£ the triangle ABC, the squares, AD and BF, are constructed, AF and DB are equal. Ex. 25. If the angle included between a tangent and a secant is half a right angle, and the tangent equals the radius, the secant y Google REVIEW EXERCIHES IN PLANE GEOMETRY 501 Ex. 26. The sum of the areas of two circles is 20 sq. yds., and the difference of their areas is 15 sq. yds. Find their radii. Ex. 27, Construct an isosceles trapeaold, given the bases and s, Ex. 28. Show thfit, i£ the alternate sides of a regular pentagon he produced to meet, the points of intersection formed are the -vertices of another regular pentagon. Ex. 29. If a poat 2 ft, fi in. high caats a shadow 1 ft, 9 in. long, liow tall is a tree which, at the same time, casts a shadow H6 ft. long I" Ex, 30. If two intersecting chords make equal angles with the diameter through their point of intersectiju, the chords are equal. Ex. 31. From a gireii point dra einal segment Is half the secant. rele whioh touehus Ex, 33. If one diagonal of a quadrilateral bisects the other diagonai, the firat diagonal divides the quadrilateral into two equi- Ex. 34. In a given square inscribe a square having a given side. Ex. 35. A fleld in the shape of an equilateral triangle contains one acre. How toaiij feet does one side contain f Ex. 36. IE perpendiculars are drawn to a given line from the ver- tices of a parallelogram, the sum of the perpendiculars from two opposite vevtiees equals the sum of the other two perpendiculars. Ex, 37. Any two altitudes of a triangle are reolproeally propor- tional to tbe bases on which they stand. Ex. 38. Construct a triangle equivalent to a given tj-iangle and having two given aides. Ex. 39. The apothem of a regular hexagon is 20, Find the area ot the inscribed and circumseribBd circles. Ex. 40, M is the midpoint of the hypotenase A£ of e, right tri- angle ABC. Prove 8 ilC~ = jJs"+llC~+Tc''-, y Google 50'-i C.KOMF/l'llY. ArrHNlHX Ex. 41. Traiififonn a civon tvianjjlo into an e angle eontaiuijig a given acute angle. Ex. 42. The area of a square inscribed in a aiea of the square inscribed ia the oirole as 2 : 'i. Ex. 43. If, on a diameter of the oirele 0, OA- aliel to BD, the chord CD is perpendicular to JC, Bi. 4-5. State and prova the converse of Prop. XXI, Bonk III. Ejt. 46. If, in a giren trapezoid, one base is three times the otlifc base, the segments of each diagonal are as 1 : .t. Ex. 47. If two Bides of a triangle are 6 and 12 and the anglt- included by them is liil°, find the length of tlie other side. Also lind this when the iuciMdod angle ia 4."!° ; also, when ll!0^. Ex. 48. How many aides has a polygon in whieh the sum of the interior angles exceeds the siita of the exterior angles by 540"? Ex. 49. If the four sides of a quadrilateral are the diameter of a oirele, the tvo tangents at its extremities, and a tangent at any other point, the area of tlie quadrilateral equals one-half the pcoduet of the diameter by the side opposite it in tlie quadrila-teral. Ex, 50. An equilateral triangle and a regular he.iragon have the Bsrae perimeter; End the ratio of their areas. Ex. 51. To a circle whose radius is 30 e from a point 21 dm. from the c Ex. 52. If two opposite sides ot a quadrilateral are equal, and the angles which they make with a third side are equal, the quad- rilateral is a trapezoid. Ex. 53. If two circles are tangent ostornally and two parallel fliametera are drawn, one in each circle, a pair of opposite extremities of the two diameters aud the point of contact are eolllnear. Ex. 54. If, iu the triangle ABC. the line AD is perpendicular to BD. the bisector of the angle II, a line through V parallel to BC bisects AC. y Google EKVIFAV EXERCISES IN PLANE GEOMXTRY i)0-i Ex. 55. BLaocl a given triangle by a line parallel to a given liae. Ei. 56. It two parallelograms have au angle of one e-^uul to the Eupplement ot an angla of the other, their areas ai'o to each other aa the produetB of the sidoa inoluding the angles. Ex. 57, The sum of the medians of a triangle is less than the perimeter, and greater than half the perimeter. EiC. 58. If PARIS is a secant to a <>,irele through the center O, FT a tangent, and Tit perpendicular to Pli, then I'.l ; l'lt = I'0 : PIS. Ex. 59. Two concentric circles have radii of 17 and 15. Piud the length of the ehord ot the larger which la taugtrnt to the smaller. Ex.60. Onthe egure, p, 244, (o) Find two pairs of similar triangles; (6) Find two dotted lines whieli are pei'pendiculac to each other; (e) Diaeover a theorem oout-erning points, not connected by lines on the figure, which are eollinear; {(J) Discover a theorem concerning sqnares on given lines. Ex. 61. One of the legs, AC, of an isosceles triangle is produced through the vertex, C, to the point F, and F is joined with I), the mid- point of the base AB. I)!'' intersects IlC in £. Prove that Cf ia greater than CE, Es. 62. The line of centers of two circles intersects their common external tangent at P. PAliCI) is a secant intersecting one of the two circles at A and U and the other at C and D. Prove PA X PD = PEXPC. Ex. 63. Trisect a given parallelogram by lines drawn throut'h a given vertex, Ex. 64. Find the area of a triangle the sides of which are tho chord of an arc of 120° in a circle whose radius is 1 ; the chord of an arc of UO" in a circle whose radius is 2; and the chord of an arc of 60° in a circle whose radius ia 3. Ex. 66. Two circles intersect at P and V. The chord CQ is t, gent to the circle QPIl at Q. APB is auy ehord through /'. Pn that AC is parallel to Bq, y Google Ex. 67. la tho trianglo ABC, from 7», tlio midpoint of Jli; J and DF are firiiwn, biseetlng tbe angles ATJIl and ADC, and maeli AB at E and AC nt F. I'rove £F || ISC. Es. 68. Produce the side KC of the triangle ABC to a point />, that PBXfC=I^j'. Ex. 69. lu a giveu circle inscribe & reetan^le similar to a f;i\ rectangle. Ei. 70. In a given sonii .cin'lo i ns.:illji) a rtcliinjjiii similiir to ?i given rectcngie.' Ex. 71. The area of an i soSL'oles trapcKoid is 140 aq. ft., one bas« la ^6 ft., and the legs make < in angle or 4:,- Willi tlic other liaae. Find the other base. Ex. 72. Cut off one-thli rd tho a rea of a given trinngie by i, iin^ perpendicular to one side. Ex. 73. Fiod tbe sidea of a tri^ ingl^ wLo-^o area is 1 sq. ft., if the sides ave in the ratio 2 : : :i : 4. Ex. 74. Divide a given line into two parts such that the aum of the squares of the two parts shall be a minimum. Ex, 75. If, from any point in tbe base of a triangle, lines are drawn parallel to tbe sides, find the loeus of the center of tbe paral- lelogram ao formed. Ea. 76. Three siiios of 'a quadrilateral are 845, 613, filO. and the fourth side is pei'pendieular to the siijes H45 and HIO. Find the area. Ex. 77. It BP bisects the angle ABC, and BP bisects the angle CHA, prove that angle F=i sum of angles A and C. Ex. 78. Two circles intersect at F and Q. Through a poiut A m one oireumferenoe lines Al'C and AQD are drawn, meeting the other in C and 1). Pr gent at A parallel ta CD. Ex. 79. In a given triangle, draw a line parallel to tbe base and terminated by tho aides so that it shall bu a mean jn'oportional be- y Google REVIEW EXEIiCISES IN PLANE GEOMETRy 505 Ex 80. Find the angle iasoribed in a asmioirole the Bum of whose rides is a maximum, Ex. 81. The bases of a trapezoid are IGO and 120, and the alti- tude liO. Find the dimensions of two equiTalent trapeioids into which the given trapezoid is divided by a line parallel to the base. Ex. 82. If the diameter of a given circle be divided into any two segraente, and a seraioireumfereQce be described on the two segments on opposite aides of the diameter, tlie ai'ea of tlie civcle will bo di- vided by the semieii'cumferencea thus drawn into two pai'ts having tlie same ratio as the segments of the diameter. Ex. 33. On a given straight line, AB, two segments of oiroiea are drawn, ^PJSnml AQH, Tlieanglea QAI' tmiQBF are bisected by Unas meeting in E, Prove that the auglo £ is a constant, wherever P aiiJ Q may be on their arcs. Et. 84. On the side AB of the triangle ABC, aa diameter, a cir- cle is described. AT is a diameter parallel to BC. Show that EH bisects the angle ABC. Ex, 85. Construct a trapezoid, given the bases, one diagonal, and an anfjlo included by the diagonals. Ex. 86. It, through any point in the common chord of two inter- aectinf; circles, two obonls be drawn, one in each circle, through the four extremitiee of the two chords a ciruumference may bo passed. Ex. 87. Prom a given point as center dsserihe a circle cutting a given straight line in two points, so that the produot of the distances of the points from a given point in the line may equal the square of a given line segment. Ex. 88. AB is any chord in a given circle, P any point on the cireumference, P\[ is perpendicular to AB and is produced to meet Iha circle at Q; AN is drawn perpendieular to the tangent at P. Prove the triangles A'AM and PAQ similar. Ex. 89. If two circles ABCD and EBCF intersect in B and a and have common exterior tangents AE and D-F aut by BC produced at <j audi/, thi^u VH'- = Ba--\-AE-. y Google aUh GKOMKTKY. Al'fENiilX EXERCISES. CROUP BT RKVIi;\V EXERC'lfiES IN SOl.lD liEOMETRY Ex. 1. A sepment of a straiRht line oblique to a piane is greatei- than its projeclioti on Ihe plane. Ex. 2. Two tetvahedroiiri are aimilar if a dihedral angle of oim equals a diliedral angle of the other, and the faeea forming tboso difieilral angles are similar each to each. Ex. 3. A plane and a straight line, both of wliic'h are parallel tc. the same line, are parallel to each other. Ex. ,4. It tlie diagonal of one taue of a cube is 10 iurlits, find Iht, volume of the cube. Ex. 5. Construct a spherical triangle o ■n a ijiv lea of the sides of the triangle. Ex. 6. Given AJ! 1 MX, Jfand KF I. MR; prove EF X I'M. Ex. 7, The diagonals of a reetangula r paral- Ex. 8. What portion of the surface of a sphere is a triangle eseli of whose angles is 140"? Ex. 9. Through a given point pass a plane parallel to two given straight lines. Ex. 10. Show that the lateral area of a cylinder of revolution is equivalent to a circle whose radius is a mean proportional between the altitude of the cylinder and the diameter ol its base. Ex. 11. The volumes of polyhedrons circumscribed about equal spheres are to each other as the surfaces of the polyhedrons. Ex. 12. Find Sand T of a regular square pyramid sn edge of whose base is 14 dm., and whose lateral edge is 250 cm. Ex. 13, It two lines are parallel and a plane be passed through each hese pianes is parallel to the given lines. y Google EEVIEW EXERCISES IN SOLID GEOMETRY Ex. 14. Givoii Pll X plane AD, Z rEU= Z FFH ; prove lI'El'=Z.rFE. El. 15. If a plniie be passed through the midpoiiila of three edges of a pnrallelopiped wliiirh meet at a vertex, the pyramid thna formed 13 what purt of the parallelopiped f a of its dist.anp«s Es. 17. Given the points A, Jl, C, I> in a plane and Pa point outside the plane, J/1 perpendieuldr \o the phme PBD, and AC p.T- pendiKular to tlie plane i'VIi; prove lliiit I'D Is perpeudieuhir tu the plane Al'.CIi. Ex. 18. In n spherb whoso radius is ">, find the aifii of a /nnti tiju radii of whose upper and lon-er hases are -T and 4. Ex. 19. Two cylinders of revolution havK equal lateral aroaK, Show that their volumes are as It : It'. Ex. 20. The midpoints o£ two opposite sides ot a quadrilateral in space, and the midpoints of its diagonals, are tlie vertioes ot pi parallelogram. Ex. 21. How many feet oE two-ineh planlt are necessary to eon- Btrucf. a box twice as wide as deep and twice as long aa wide (ou the inside}, and to contain 216 on. ft.! Ex. 22. If two spheres with radii K and r are concentrie, find the nreu of the section of the larjjer sphei'e made by a plane tangent to the smaller sphere. Ex. 23. In the frustum of a regular square pyramid, the (!.!;.■■■» of the bases are denoted by b\ and hi and the altitude by H ; pvusu that X=iv'(6i— i-il'-h^fl^ Ex. 24. If the opposite slJ V .>|..posite angles are equal. y Google 508 (jEOiii'iTKY . APPi;>; dix Ex. 25. Obtiiin the simplest formiita for the lateral suvfaee of a truDeated tvianfrular right prnm, each edge of whosf base is «, and whose lateral edges are 21, 9, and r. Ex. 26. The area of a zone of one ba^o is a mean proportional between the reiuainiiig surface of the sphere and its eiitiio surface. Find the altitude of the zone. Ex. 27. The lateral edges of two similar frusta are as 1 ; <i. How do tiieir areas compare ? Tlieir volumes f Ex. 28. Construct a spherical surface with a given radius, r, which shall be tangent to a given plane, and to a given sphere, and also pasa through a given point. Ex, 29. The volume of a right circular cylinder equals the area of the generatiuf- rectangle multiplied by the ciroumferenee generated by the point of intersection of its diagonals. Ex. 30. On a sphere whose radius Is Si inches, find the area of a zone generated by a pair of compasses whose points are 5 inches Ex. 31. The perpendicular to a given plane from the point where tlie altitudes of a regular tetrahedron intersect equals one-fourth tho sum of the perpendiculars from the veitieee of the tetrahedron to the same plane. Ex, 32. Two trihedral angles are equal or syminetrieal if their corresponding dihedral angles are equal. Ex. 33. On a sphere whose radius is a, a zone has equal banes and the sum of the bases equals the area of the zone. Fldd the alti- tude of the zone. edges of a tetrahedron Ex. 35. Find the loeus of all points in space which have their istances from two given parallel lines in a given ratio. Ex, 36. It a, b, c are the sides of a spherical triangle, a', 6', c* iie sides of its polar triangle, and a>b>c, then a'<l/<e'. Ex. 37. A cone of revolution has a lateral area of 4 sq, yd. and n altitude of 2 ft. How much of tho altitude must be cut oft by a ilane parallel to the base, in order to leave a frustum whose lateral y Google EEVIEW EXERCISES IN SOLID GEOMETKY Ex. 38. The total a isoribed sphere as 9 : ' Ex. 39. CoQstrucl a sphere of give be thngent to three given, spheres. eu. in. and its altitude ia 20 in. Show how to find the radii of the baaes. Ex. 41. On each base of a cylinder of revolution a cone is placed, with its vertex at the center of the opposite base. Find the radius of the circle of intersection of the two oonieai surfaces, Ex. 42. The volume of a frustum of a cone of revolution equals the sum of q cylinder Bud a eone of the same altitude as the frustum, and with radii which are respectiveiy the half sum and the half differ- ence of the radii of the frustum. Ex. 43. A square whose aide is a revolves about a line through (mo of its vertices and parallel to a diagonal, as axis ; find the surface and volume generated. Ex. 44. If a cone of revolution, roll on another fixed cone of revo- lution so that their vertices coincide, find the kind of surface gen- erated by the axis of the roiling eone. Ex. 45. An equilateral triangle whose side is a revolvea about au altitude as au axis; find the surface and volume generated by the inscribed circle, and also by the eireumseribed circle, Ex. 46. Find the locus of the center of a sphere which la tan- gent to three given planes. Ex. 47. If an equilateral triangle whose side is a be rotated about a line through one vertex and parallel to the opposite side, as an axis, find the surface and volume generated. Ex. 48. What other formulas of solid geometry may be regarded as special eases of the formula for the volume of a prismatoid ? Ex. 49. Through a given point pass a plme which shall bisect the volume of a given tetrahedron. Ex. BO. In an equilateral < leuts are porpetidicuhir at tl y Google PRACTICAL APPLICATlOnS OF PLANE GEOMETRY EXERCISES. GROUP S8 (Book I) 1 Take 1 piece of pipir li urns i '^IiiirI o iaim I !i[;lit aiit,k \\ hjt gtomi I ( liGO inil fol 1 thp paper I |iriiKi]ik en definition Wh^t RComHru .i.nuplK hu ^ou nU)cout'iii!eanp:l halt the anfrle, j: thp dijgruni (tL ..f thP c llPt«Cl note the ■upp the iteis out bye l^e It IS csMutull^ the mrfhod wfd pe ind hcn(u m corn ctijig inslni itiaiuhL pcli,!, test the j,ipuiii\ of the I iiippJifLi^ -jqiiri ljy u meth'xl he tliiigr nil Ilo« thtn would ipLUi in of tin 111-kU uirIp of 3 InE\ iproiethit lh( PI square if there he auj cqii il« o side lines of the square as sho«n ii and show that e + r = j;) T-his prmeiple la irapoitant btea in correcting the a\is of i teloioc menta of which thp t Icicopt h a part ; HUrieving and astronomical instruments 4 By use of a ciipenters square am sltaight edge lay off i strits of pnnllel lines \\hat property of pii lilel line^ ba^e joii used' 5 Tell how to (,onBtru.ct a carpcntoi i miter box 6 Ihe (ii^run hhows a di miiiK instr imuit edled thepirill 1 lulir, lh( dotud outlm of thi. in-.tiuni(nt in another jiosilion th pirt RS leimining faxiil Thp dist mci i'Q = hS, PR = Q'^ HeiiLL show thit PQ ii patalltl to R!> In like manner show thU PQ i» imralld to RS Hi nee Awv. PQ i^ panllLl to P Q If \ line b( ilriwn p( rp< nditular to PQ and anothrt p(ip<ndicuUr to P Q the pctpcndiiul ir ll u~ d i pmllci to lach i.thir { \it 111 lin,-, i Liptu hi. In to i , ill Uii i. art, parjlkl) 510 y Google PHACTECAL APPLIPATIOXS f)!! The above are ctbcs of linked motion, a kind it mMhanism o[ mdo importance Observe fonnatance the system of hnks which oonntLt the driving wheels ot j loLomotivi nith the [iiaton in the tyhnder and also the jointed rods conntctiug thi walking bi im ot j, atearaboa! nith the engine Look up ilso, m the Cmliuv DictionavVi for instuncL the words hnkagt, icii, and jiaialkl motion 7. The distdnec- bpifliiii paflaable bainer (h» b a pond) may be Idi i no instrument for in i Ut A and B Ix I hi > iMit Bliitioii ( a,nd meiaiirL to F, making ( I = l< DC = liC Mtasiirt tJl 1! AC = Cr -220 ft,; liC - long is AB? 8. In the trusses of steel bridges, why are the beams nnd rods arrai^ed so as to form a network of triangles as far as possible, and not ot quadrilaterala, or pentagons, for instance? (See Ex. 3, p. 70; also Art. 101.) How 13 this principle also made useotin forming the frame of a wooden hiiX'^Q oiboxcir or tostren^thni i « eak f i ime < r f ener of an\ kind' 9 Draw a map for the lollowmg sar\(j Hot's to the seilo ot 400 ft to lh( inch In Hjitir! off the an^k'i dia" i. dott(.d lorth and south line through a<!i stitioti ml then use a l^^o iCHs-ii>li ph(e, sepaiated bi in m tlioi 1 luml L l( iiadHI I'lodi I'rodULG m U. I> 1 ;;,:■;. .^^. ' " / \ ,1 1,11 u. Phul lis = 1)1 liC = Cl> -liiOtL; ami IJF - 21(1 fl., 1 Stations |„, iiing=i Di,l mceh 1 JN "0 i m> fl 1> ' \ J)i) fl 1 s r"} JiOfi ' D 1 ■^ .o°\\ bOOlt Kttp your dialing £or a iltr iiw 10 Obtain oi nnkeupawLt ot Bur\e\ nolcf, snnd ir lo those in E\ md niakt a lii using tor them U IitD( and/Tfp ol2l be tno naih iRiiMndiiiilar to the ]>Iiti( fthtpipti Lit smiUmnrorslji' itluhi-d to those » Ulh at land/) Ul Z( - W I,,turi\ of hght pi«i Ihiough y 1o 1 bircflul I I 1 /i inclthimoto/' ProvctSiil/ l/'B is i right ingle [->LC The Ian of reflicted light is that the ai^le of inLideni e e(]n i\- thi angit of reflection or in the figure x = x and u = y ili(.u Jit thi triinuln ABC j; + ,j + i^" - ISO'' or i + y = n-i° \l out thi p, iitB 1 ani b it i-'t +21/ + 1 = 100° u + d = tJO ttc ] y Google (";eometry Bht c gf t s Kbt Rl 1 II h 1 113 t gl thth ray coming from Q through /' to A. IS, The velocity of light is dolermined by the ' " in till! following manner: J_ pI.TJie of the paper ami livot; let OP, bel,-l,fi;. it and reflected Ihroujih ills distant, whence it i> retuni of fhe ray to 0, a rotating Let A,Bi be a mirror rotating about as u ] Let to be a ray of light striking the miritir : M to a small Rtafionnry mirror some m ref)i:ttcd back through .1/ to 0. On the A,B, will have rotated through a small angle, o, to the position AiB-; hence the ray will be reflected in the direction OR, Let the pupi! show that Za = I^Z LOR. Since Z. WR may readily be measured, ^ a is known, and if the rate at which the mirror is rotat- ing is known, the time occupied by the ray in traveling from to the stationary [Suo, ^PiOPj = Z /l.O^! (Art. 132) Z_P,OL = Z. PvOM = cc ( / incidence = Zreflcfliuti) .■- Z P^OL = x~a .■. ZiOfi =» + a- (x-a) =2al If Z LOR = 2° 19', OM = 3 mi., and the mirror AB rotates 100 times per second, determine the velocityof light per second. 13. To prove that the image of a point in a plane mirror is on a perpendicular from the point to the mirror and as far behind t!io mirror as the object is in front of it, let MM' be the mirror and P the point, and ./> PAE and PA'E' two reflected rays. Show and back Js determined. Z PAM = AEAW = £ P'AM. :. Z /M.U' = Z P'AM'. . PAA' = 4 P'AA'. .-. PA = P'.i, etc. 14. From the above show the direction in which a billiard ball must be sent in order to strike a certain point on the table afler striking oat side uf the table. y Google p h f \ 7 / b / -' PRAmCAL VPPLI'^ATION'* ■51'i lors 0.1/ anil 01/ bp peiprnd :hr to pt h (thpr makp i pon'if ruction to show nhere the point P mtil appear to be to an e>e at E aftir tbe light from P hia hppn iifltctpd from both mirrors hoc Pio\eSP'' = A 1 + 4P = B 1 + IP ] 16 ■\\hat woull liL thL ippU- i ition of F^ 15 to a ball -truck jn a biUiaid tabU 17 The path of a ni} of light before and iiftcr enlcnnji; a Rlaia prism IS gnen b\ ihc Imes AB and CD The entire angle bj «hich a raj of light IS defieetpd on passing thro igh the prism is denoted by x Prove that J- = , + r - P (Zi T'Bn and PC^aie 1 1 ZO ta. By use of d quadrilateral BPCy, y = 180° - P. Then use the quadrilateral all of whose sides are dotted lines,] EXERCISES. GROUP i 2, By use of the c.irpenfer's sqiiaro find the center of a given circle. 3. Show how a. pattern maker by the use of a carpenter's square can determine whether the ciivity made in the edge of a boatd or piece of metal is a Beniicirclo, 4. Biboct a (civcn angle by the use of a carjientpr's sqiiaic. fi. By use of squared paper divide a line U inches long into five equal parts. Info 7 equai parts. Into 3 equal parts. Can you make this division on paper ruled in only one direction? 6. Make tip ami work a similar exatjiple tor yoursolf. y Google (GEOMETRY long and let be ifK midpoitiL Murk off four equal parts at D, F, li. With O through C, Z), P, H. A. From draw radii (one of which is OA) dividing the angular space about into eight equal pxrts Beginning at dran tht smooth curve ( VHi-tb through thp joints where the ladii drawn inter ect the circumferpin.es ui, indieatcil m the diagrim Ihc lesult will be the outline of i special foim of wheel (ailed a C'jm much UM'd in machine t mo- th g aid ] ea p t b Pe g 1 a; 1 l g hoes A ; 1 t t back and forth motion. TI i ff f 1 g h bet on the diagi'am ia termed tl ti f th 8. Make a lirawing of } h th thi h the longest radius IJ in. 9. Stretch a 100 ft. tape or part of it so ai 10. To extend a straight line AB beyond an obslatic, ;is a building, we may proceed as follows: At B measure off an angle C ,. ABC = 60°. Produce CB to D, /\ /\ \ etc. Let. the pupil complete the a — '— ^^ /^ — ^ — —K construction and prove lliat his \ / method is correct. V 11. Let the pupil solve the problem of Ex. 10 by the construction of right angles instead of angles of 60°. 12. To find the distance between two points A and B, one of which A l"1 ^■^^ '^ inaccessible, we may proceed as fol- p _ u--_j,B idYjg: Extend BA to C. Measure a con- Q/^''%\ venient line AD and extend AD to E, ^^^-^y f| making DB = ^D. From £ run a line 1| .F^ E \\. ■^'' ^""^ meeting BD extended at F. ^ Measure £f, Prove that £/■ = A£. y Google PRArXICAL APPLICATIONS 515 if' 13 Show how to solve the problpni of F\ 12 iij' c-onsfriieting a hne at right angles to xnc Iher hne instead of one parallel to inotlier, 14 If two Eirfetfi meet as in the diagram show hoB, a cur\o of gi\en radius r may be made to tike the ylaee of the angle 4 and be • tangent with the curb of the two btveets 16 The curve in a railroad track is usually ao arc of a circle tangent to each straight track T which it joinn If two straight fra^-l « 4B and r>. '"D oto joined bj a circuUr curve tangent to I ofh Dt them and P la the point where IS and ( D \o !J mft it e\tKnded piove ZTI D = 2 Z. Tii 16. One na> of Ujmg off a railroad curve n track is as follows; Let ABP be the ^ given straight track. At B construct a small angle PBC (whose size depends on the degree of curvature which the curve is to havet. On it mark off BC = 100 ft. y Construct Z CBD = Z P^C. Take C as a center and 7 100 ft. as a radius, and describe an arc cutting BD / at D. Construct E from D in like manner. Prove / that, B, C, D, B all lie on / the arc of a circumference tangent at B. [Sua. Pass a circumference through B, C, D. Prove that B lies on this circumfer- ^e and that ABP is tangent to it.] 17. Let AB and CD be two straight rail- O road tracks connected by an arc AF of a circle whose center is 0, and an aic FC whose center is 0', the arcs having a com- mon tangent at F. Prove that the angle of n (z) of the two straight tracks, if produced, equals the sum of the central angles of the two tracks. [Sue. X ~y + z. Use Ex. 15-1 A curve like AFC composed of two or more arcs of different radii is called a compound curoe. Can you suggest why a compound curve should lie used in connecting railroad trajjks? 18. If Hip two arcs which compose a compound curve lie on opposite sides of .their common tau^i;iil, the comjiounii curve ,ia called a reoersa ) AB y Google GEOMETRY Thus on the diagram, BCD nnecting the itraight roads AB and DE. Discover the relation between \% Whiiij is ii railroad Irogl' If a ciirvod track c track, show that the angle of the frog (i) equals the central angle of the curved track (o). 20. If two tracks which curve in the same direction crosB each other, find the relation between the anglo . o[ the frog and the central angk-s of the two curved tracks. 21. Find the same when the two curved tracks curve in opposite dii'ontions. 22. Show how to locate a gas-generating plant (P) along a given j^ straight road AB so that the length of pipe / connecting it with two towns (C and D) "■■v^ / shall be a minimum. "'-./ , (Use Ex, 24, p. 176.) 23, Discover and state ;; Ex. 24, p. 17U, similar to one given in Ex. 22. 24. If a treasure has been buried 100 ft. from a cert; distant from a given straight road and a path parallel t( how the treasure may be found. 26. Make up and work a similar example for yourself using the prin- ciple of E)i. 2, p. 167, Also using that of one of Exs. 3-S, p. 168. 26. Prove that the latitude of a place ^n the earth's surface equals the elevation of the pole (that is, on the diagram, prove Z. OE-4 = Z.PAO). „ Z_ 27. Given the sun's declina- tion (i.e. distance north or south of the celestial equator), show how to determine the latitude ^ ^ ' of a place by measuring the zenith distance of the sua- Also by measuring the altitude of tb.e sun above the boiison. application of o the road, show y Google PRACTICAL APPLICATIONS 517 3 and 27 i 28, How was Peary aided by the principles of Eks. determining whether he had arrived at the North Pole? 29. If on April 6 (the day of the year on which Peary was at the North Pole) the sun was 6" 7' north of the celestial equator, how high above the horizon should the sun have been as observed by Peary? At what hour of the day was this? 80. The sextant is used almost daily by every navigator in deter- mining the altitude of the sun above the horizon, and hence the latitude of the ship. The construction of the sextant ia ^ based on the principle that if a ray of l^t be reflected from two mirrors' in succession, the change in the direction of the ray of light equala twice the angle made by the mirrors. Prove this law. |Su(j. Let M and N be the mirrors and the reflected ray be SMNO. It is required to prove that Z.y = 2 ^x. It is to be noted that Z.FNM, being an exterior angle of triangle NS'M, equals x + i. Hence angle MNO equals \ / 180° -2 (e -l-t). Hence in triangle MNO, \^/ y + ISO" - 2 (a; + t) + 2 1 = 180°, etc.] \^. In the sextant, angle y is the elevation of thi" sun above the horizon, and x can be read on the rim of the u 1 position V mirror N i» fij.ed i M rotates It maj be observed thit the prin fjple of this example is the ^an < la that pro\ed m E\ 12 p 512 31. The diagiim show? a some- what simple aubstitutp tor the e\ tant called the angle meter. ih the center of the arc BC and MM' is a fixed mirror. The instrument is held BO that a ray SO from the sun when reflected from the mirror passes into the eye (A) in a horizontal direc- tion. Show that the elevation of tliE> The rim BC is graduated so (hat y Google ol8 GEOMETRY The angle-meter may be used to angles. What is the advantage in using a mirror on sort? That is, why is not the positioD of the across a graduated arc? 32. Draw an easement & liorizontul as well a; instrument of this n observed directly i composing the e (.AB) tangent to the rake cornice Bc and passing through a required point A. (The e construction is used in layiii^ out the ease- ments of stair rails, etc.) (Use Ex. 7, p. 174.) 33. A s^mental arch is a coraijound curve composed of the arcs ot thriie circUs. The method nof constructing a segmental arch is as follows; Let AB be the span and CD the altitude of the re- quired arcli. Complete the rectangle GADC. Draw the diagonaJ AC. Bisect the angles GAC and GCA. Let the bisectors meet at E. Draw EH perpendicular to AC, meetii^ AB at JV <md CD produced at U. Make DK equal to '^' DN. Then show that N is the center and A'.'l the radius, H the center and HE the radius, and K the center and KB the radius for the a arch. (See Hanstein's Constructive Drawing.) [ScG. At E draw a hne perpendicular to EH. Then Z. -VB-1 - Z. NAE. (Complements of equal angles are equal) etc.] 34. What is called a Persian arch may be constructed as follows; Let AB be the span and CD the altitude of the required arch. Draw p n P *he isosceles triangles ADB. Divide AD into three equal parts at H and G. Construct HK ± AG at H, and meeting AB produced at K. Produce KG to meet Ef which has been drawn through D]| AB. With K as & center and KA as a radius, and . B as a center and EG as a radius, describe arcs meeting at G. Prove that these area have a com- mon tangent at 0, and therefore form a compound 5urve. (See Hanstein's Constructive Drawing.) 3B. Construct a Persian arch in which the arc DG = arc NG. [Soa. Bisect line AD instead of trisecting it.] 36, Construct Persian arches in whicli llio chord:; NG and DG have y Google :ri PRACTICAL APPLICATIONS 519 various ratios, and decide wliich of these arches you think is the most beautiful. 37. Construct aegmental arches of various shapes and decide which of these you think is the most beautiful. 38. Construct a trefoil, given the radius of one of its circles. [SoG, Construct an equilateral triangle, one of whose sides equals twice the given radius. Take each vertex of Ihe triangle as a center, and half of one of the sides iw a radiua and describe arcs.l f ^ 39. Inscribe a trefoil in a given equilateral triangle, V_yV^_y [S0G. Bisect, the angles of the tiiu,ngle, etc.] 40. Discover the method of eoiislrucdou of each of tlu' figuvpr! or diagrams on the next page, and then rcpi-oduce each <•( tiicm in a drawing. Squared paper may be used to advantage as an aid in making some of these constructions. See Fig. 10, p, 520. 41. Construct a diagram siiailar to Fig. 10, but making a rectangle instead of e, square the basis of the drawing. 42. .\lso one making the rhomboid the basis. 43. Construct a trefoil and develop into !Ui ornamental design by placing small circles and ares within its parts and larger circles outside. 44. By use of squared paper invent designs similar to Fig, 10 on p. ,')20, but formed by two intertwining lines, 46, Collect a number of pictures of ornamental designs, tracery, scroll work, etc., such as arc vised in architecture, wall paper, and similar patterns, whose construction depends on the principles of Book II. Show how these designs are constructed, and reproduce them in drawings. EXERCISES. GROUP 90 (Book III) 1. To tind the distance between two accessible objcpt.-i ;1 and B separaliid by a bari'iet', take a convenient point C, measure AC and BC. Protluce EC to I) and AC to F so that DC = some frac- tion, as i, of CB, and FC = the same traction of FD. How, then, is All com- y Google rmre y Google PUACTiCAL APPLICATIONS 521 Let BC = 300 yd., DC = 60 yd, FD = 52 yd., find .IB. Why is this method of finding AB of Ex. 7, p. 5U. 2. What similar method doea the gram surest for finding AB? 3. In the di.igriim of Ex. 2, between A and li, and ako between nieaaurements would be iieeetaary of AB> Of BF' i Mork lUinCx 2 p 310 6 In CJ.S ihi un i nnt '■hinm' , AC = 240 yd., FC = 4S yd., often more convenient than that adjoining dia- of the top tf 1 1 fiom the groii il h 1 ci cr tai the mirror l. 1 '0 ft fi im the tree how h gh la th tree' 7 I rcsters often detei mine the ht kH of -v tie e by ^n n stiument ill i raitetmms Hefei It Meaa- m IhMi m,,lean^ hich tins initiumtil 1 un liiieted n ohoun in thi, di Jijr i n If the i -Ian e fiom 4 t ) the foot ofthotieei bOlf h( = J iiichts mdLF =4 n ht■^ hi 1 1 e 1 tight of the tree suit n fron -1 roimF \ad shadons cannot be used the heisht of an objett like a tiet, or stteplc cin often be dctLrmined bj a method in dicated m the d r i w i n g W hat di'itanee mutt be meas lied and whv to determine thi, hught of the tret* 6 'ihow how to find the h „ht of I tret by placing I 1 irior m a honzontal posi- tion on the ground and stand- ing fco as to see the reflection if the oijserver's eye is 5i ft. 1 t , the y Google 522 r.EOMKTIiV ACB. On paper make adiaiving of llio triangle ACB to a convenient scale. On lliis drawing mraanrp l.iic Wm. wliifh represents AB and hence determine the Icnj^lli of AH. By use of this method ivc arr> riiivrd Ihn labor of marking out and measuring the lines CD, CF, and DF. Apply this rtiethod to the raeasurcraenfc of two objects in your noi^i;h- borhood which are aepai'ated by an impassable barrier. In like manner show how the distance from a given point to another inaecessible place may be dptermined. 9. Also the distance between two place." both of which are inacces- 10. What is the meanuring inst.rument called the diagonal Boale? How is the principle of similar trianglca tised in it? Show how tile diagonal seale aiils in the accurate moasurement of lines and hence in the accurate determination of a distance such as AB in Es. 7, p. 511. n 11, Id the triangle OAB, A is a right angle, and 0.1 is 1. By a method which is Ijeyond the scope of this book, the length of AB is computed and found to be .839 +. Using this tact find the second triangle, of the accompanying table, find RQ if angle P is 10°. 20", Tables ^ving the other sides of all possible right triangles when one side is unity have been computed and when used as in Exe. 11 and 12, form the basis of the subjwt of Trigo- nometry, ity use of this science, after measuring the length of a single lino a few miles long on the earth's surface, we an d t m*n th i' ( n es and laf po9 1 ons f th pi i-p^ th usands f mis ana w ihout a. g an t en ng 1 n By u f tl e« esuU as a bas th d tan e f th moon dete m aed aa ipp o\in t t 40 000 miles of the ) SOOOOO n le- and of th n -e t fa d s( ''0 000 000 Omi 01)0 I Anglo AB 08 10 ,176 I.OIS .364 1.064 .Till 1.155 40 .839 1.305 )0 1,192 1,556 ()0 1.732 2.000 70 2.747 2.924 5.671 5.759 y Google PR\rTIC\L APPLICATIONS A knowledge of these diotances ha^ led to importinl impioi.enii in method'5 of iixMg,ition and havi, thus fanihtilt^ travel ^nd merce and increased their benefi''! foe ( usaU 13 On thpdiagramgi\enAJSif,i)Onoii and the angles as indicated, compute the length of CD, by use of the table given in Ex. 13, 14 To the diagram in Ex 13 annex another tnangle CFD giving it angles %vhi h are multiples of 10° and compute ll length of CF 16 A yanloqraph is an in trument fui drawing 1 phne figun iinuhr to i rimii plane fagurt It is uaetl for enlai^mg oi -^ uuwu. i> reducing maps and dra^vmgs It consist-, of four bars, parallel in pairs and jomted iX c b C and E as shown in the diagram, <i)EC IS a parallelogram The rods may be joined so that any required CB' fixed pivot and pencils are at b ind F. Show that A, b, and B are equals the given ratio A turns upon a carried [Sua, Dra^v a line from .4. to b, and a line from A to B. Prove the triangle Acb and ACB similar {Art lt^3); henw show that Ab and AB coincide, etc.] 16 Using the fact that a triangle whose sides are 3 4, and ^ units of length is a right triangle show how by Btretchmg a 100-ft tape, U> ponRtruet i right angle is accuiately is posi^ibte (\mong the ancient Egyptians a class of workmen existed called rope stretchers, whose busine** was to construct right angleb in this genera! w i\ ) 17 The sttongest beam which can be cut from j gnen round log is found as follows Take AB s diameter of ihi log and trisect it at C and D Diaw C E and Dt ± \b n\ meeting the iucumf<rimt \\ h a.tidFrp«puli\(K Draw AF FB BL and If Pro\e iFBF i rcdinnU also FB U - 1 v^ or iprroxinitih i > 7 y Google (324 GEOMETRY |SuQ. _£B is a mean proponional beUveen DB and AB. Therefore FB' - i AB" (Art^43).] _ In like mnniiM F-V - I AB', ote,] 18. A sphere S weighing JOO ib. rests on ihe inclined plane AB. AC contains 8 units uf length, BC 6 units. A force which prevents S from rolling down the plane would be equivalent to a lifting force of how many pounds exerted on S and pai'ailei with l/i'' [Suo. Ile=!ohe the n eight of S (represented by SP) into two lorces, one perpendicular to AB and the othei parallel to .-IB. Prove the triangles ACB and "il'B similar, and obtain AB: BC = SP: SR, or 10: 6 = 100 lb SR] The principle involved in this example is of great practical impor- tance. Thus in many maoliines useful results are often obtained by representing a force by the diagonal of a rectangle (or parallelogram) and separating this force into two component forct* represented by the sides of the rectangle (or parallelogram), onlj one of these com- ponents being effective. This principle makes poa>ib!e (he action of the propeller of an aeroplane or steamboat, of the best wat«r wheels and wind mills, and indeed of all turbine wheels. It also determines the lifting power of the planes of an aeroplane. 18. A wagon weighing 1800 lb. stands on the side of a hiU which has a rise of IS ft. for every 100 ft. taken horizontally. Hence what force must a horse exert to keep such a wagon from running down hill, fric- tion being neglected? 20. Make up and work a similar example for yourself. 21. Show how the diameter of the earth may be determined by the following method: Drive three stakes in a level piece of gi'ound (or in a shallow piece of water) in line, each two successive stakes being a mile apart, and lot each stake project the same distance above the ground (or water). By use of a leveling instrument, determine the amount by which the middle stake projects above a horizontal line connecting the tops of the end stakes. This distance will be found to be 8 in. [Sue. Use Art. 343. .\n arc a mile long on the earth's surface may be taken as equal lo its chord. Then from the diagram of Art. 343 we obtain the following proportion, the diameter nf the earth; 1 mi. = 1 mi.; 8 in.] 22. AI?o show that the liistance that the middle stake projects above y Google PRACTICAL APPLICATIONS 525 the horizontal liiip connecting the top of the tno end stakes varies as the squai'e of the dista,nce between the end stakes. Thus if the two end stakes were placed three times as tar apart !\s in Ex. 21 [that is, 6 miles apart instead of 2 miles) the bulge of the eai'th between them would be 3' or 9 times what it was originallj-. Hence determine the projection of the middle stake (or bulge of the earth) when the end .sljikos are 4 mi. apart. Also 8 mi. 16 mi, 33 mi. 23, At the seashore an observer whose eye was 10 ft above sea level observed a distant steamboat whose hull was hidden for a height of 12 ft. above water level by the bulge of the earth. About, how far off Wiis the steamboat? 24. A seaman in a lookout 42 feet above water level with a glass could barely see the topsail of a, distant ship, and estimated this top- sail to be 45 ft. a,bovs sea level. Estimate the distance of the observed vessel from the seaman. 2E. Work again Exs. T-S, p. 310, Group 56. 36. The moon's distance from the earth's center approximately equals 60 times the earth's radius. A body failing at the earth's sur- face goes 193 in. in 1 sec. Hence, if the law of 1, the distance fhe moon falls gravitation is ;. toward the earth will be 193 ir Inth. diagram, let be the earth's center, ABE the moon's orbit, and CB or AD the distance thi- moon falls toward the earth iu 1 sec. Taking the month as 27 da, 7 hr. 43 min. 11 sec, show that .-ID 193 i n 60= approximately. This is the ealculation used by Sir Isaac Newton in testing the truth of the law of gravitation, 27, Any rectangular object, as a book, door, or photograph, is con- sidered to be of the most artistic shape when its length and breadth have the same ratio as the segments of a line divided in mean and extreme ratio (Art. 370). In accordance with this rule, if a window is 6 feet high, how wide should it be? The division of a line in extreroe and mea Golden Section. Byraany theGotdenSectio: tal principle of esthetics, having applicationi tions of the ideal humas form, in esplanatio n ratio has been termed the n is regarded as a f undamen- ! in determining the propor- n of musical har . Make up and work nn example similar to Ek, 27. y Google •a! GEOMETRY EXERCISES. GROUP 91 (Book iV) The supporting power of a wooden beam (of rectangular cross n and of given length) varies as the area of the section multiplied by the heigiit of the beam. I If the cross section of a given beam is 4° X 8°, com- e the supporting power of the beiim when it rests on the narrow edge (4") with il« supporting power when it regis on its wide edge (8'). 2. Two beiims of the same length and material have cross sections uhich are 2" X 4' and 3° X 8' respectively. Find the ratio of the supporting power of the two beams, 3. In Ex. 17, p. 533, compare the supporting power of a beam cut from a log in the manner indicated, with that of a square beam cut from the same log. 4. Also with that of a beam whose width equals i of the diameter. 5. When an irregular area Uke ABCD i calculated by means of equidistant offsets, the following rule is used; To the halt sum of the __ initial and final offsets add the sum of all the S intermediate offsets, and muUiply tiie sum by the common distance between the offsets. Prove tiiis rule. 6. Show how the rule of the preceding prob- , lem could be used to calculate an area whose re boundary is an irregular curved line. . Surveyors often determine y the area of a piece of land, as of ABCD, by taking i,' - an auxiliary line as NS, measuring the perpendicular disUnces from A, B, C, D, E to NS, and the inter- cepts on NS between these perpendiculars, and ^" combining the areas of the various trapezoids (or j triangles) formed. Supply probable numbers for the lengths of lines on the diagram, and compute the ""\ E area of ABODE. ^ , a d (, (t.. a T ^' '^f^Quently (as when the center of '~^~j~~~J: i ~ '^e curve cannot be seen from the curve)' ' K'~'--i,^^ a railroad curve is laid out by construe- ^^'^ ting a aeries of equidistant offsets per- peadicular to the tangent of the curve. mmm y Google '■---.1 ^ "ir ~--4^ ^-~-^fi! ■/ " ■"--^ i PRACTECAL APPLICATIONS 527 Thus, if ABT is a straight track and BC is a Rurve to be laid out tan- gent tfl AB at B, marli off 6ai, aioi, ajoj, all = d, and construct the _L offsets a b\ a«ft" Ojfej by Uiing the forniiiU a bu ^ r — Vi^ — n'd' n here!" IS the radiua of the curve Prove thit this formula is correct. 9 The diagram represent'" tno atiaight parallel railroad tracks connected bj a compound curve (in this ^ ^ caKe called a crosa-over tta<k) comiostd of p -r*\ t«o tracks with common tangents at D and / 1 b with centers and and having equal -i / ; ridii Taking the magnitudes as indicated ) /' ]F oil the figure show that A/ODisareif angle (use 4rts 122 IfiOl 10 Find u formula tor i m teims of ( d and w (TJae the right tnjnt,U PO. ind 4rt 401)) \lso tsr ( in tetnib of J^ 11. If i« = 4' Hi", d - 10 ft., )■ = 1,^0 ft., i/ find I. '' 12. IE a steamboat is traveling at the rate of 12 miles an hour, and a boy walks across her deck at right angles to her line of motion at the rate of 3 miles an hour, draw a diagram to show the direction ot his resultant motion. From this determine the spi;ed at. which he is going. 13. Make and work a similar example for yourself concerning a mail bag thrown from a train. 14. Also concerning a breeze blowing into a window of a moving trolley eai'. 16. Two forces, one of 300 !b., the other of 400 lb., ai-( at right angles on the same body. Find their resultant. 16. Make up and work a similar example for yourself. 17. A river is flomng at a rat« of 4.25 mi. iier hour, and a man is rowing at right angles with the current at a rate of 3.75 mi. per hour. WTiat is the result-ant velocity of the man? 18. If a star has a velocity of 15 mi. a second toward the earth and a velocity of 20 mi. a second at right angles with a line drawn from the star to the earth, find the velocity of the star in its own path. 19. A body is moving in the straight line. ID past the point (p. 528). The distance passed over in a unit of time = AB = BC = CD. When the body reaches B it is acted on by a force which impels it towaj:d y Google GEOMETRY a diBtancc BF in a unit of 1 ii !■ bv ;i line draw by til Prijvi! lluit 'iio aren swopl over in t!-. m the body tu will be iinchangc ; of thf now force. (Thai is, on ilu' .liagram prove that AOCB =0 AOPB.) This prinoipie iiccoimts for the fact that the eaiii! (or another planet or a ooiiiet) moves fuKter in i;, orbit the neoi-ei' il is l-o the Kim. 20. By traeing tho oiiliine of a map on squanvL paper, show hoiv to find the aj'pa of an irregular figure as of some oountry or pari of a country. By iiBU of this method find the area of aomc part of the state in which you live. exercises. group 92 (Book Y) 1. k cooper in fitting a head fo a barrel takes a pair of compasses and then adjusts them till, when applied six times in succession in the chine, they will exactly complete fhe eireumferonco. Hr- Ihen takes the distance between the points of the compasses as Ihe radius of the head of the barrel. Why is this? 2. Draw a square and convert it into a regidar octagon by cutting off the corners. ISuG. Draw the diagonals of the square, bisect their angles of inter- section, etc.] 3. In heating a house by a hot-air furnace the area of the cross section of the cold-air box should equal the sum o£ the areas of the pipes conducting hot air from the furnace. If a given furnace has three hot- air pipes, each 6' in diameter, and one pipe S" in diameter, and the width of the coid-air box is 1.5°, how deep should the box be? i What is the most conMinif nt Tva\ of determining the diameter of a hot air pipe if you haic no alhpers and the ends of the pipe are not accessible' 6 ^ half mde running tiack is to hait equal semicit cuUr ends and parallel straight sides The extreme length of the rectangle together with the semicircular end*! is to be 1000 fl Find the width of the rectangle [Sue Denote the length of the i \d\u> ot ihe emiciriukr ends bj I and that of one of the parallel idi. straight traj.k=i b-v y and oLtnn a pair of ■■laiultaDeoua iquafioni | y Google PRACTICAL APrLlCATlONfi 529 6, A belt runs over two wheels one ot which has a diameter of 3 ft. and the other of 6 in. If the first wheel is making 120 revolutions per minute, how many is the second wheel making. How many revolu- tions must the first wheel make in order that the second may make 300 revolutions per mtrmte? 7. If in laying a track a rail 10 ft. long is bent through an angle ot 5° 10', what is the radius of the curve? Assuming {what is not strictly true, owing to friction against the sides of the pipe, etc.) that the rate of flow through it cylindrical pipe is proportional to its area of cross section: 8, Work again Ex. 12, p. 277, 9. If a Ij-ii- i'ipG is replaced by a 1-in. pipe, how much i,-- the flow of water increased? 10. A 3-in. pipe is to be replaced by one which will deliver twice as mui;h water per minute. Find, to the nearest quarter of an inch, the diameter of the new pipe. 11. A 2-in. steam pipe conveying steam from the boiler to the radia- tors in a school building is found to supply only two thirds the needed amount of steam. What is the smallest even size of pipe that will convey the needed amount? 12. A city of 40,000 people is barely supplied with water by 12-in. mains from Ike reservoirs. If these mains are lorn out, and IS-in. mains substituted, what future population of the city is allowed for? 13. A steel bar I in, m diameter will hold up 50,000 lb. What load would a bar I in. in diameter carry? What would be the diameter of a round (cylindrical) bar to carry 150,000 lb.? 14. It Iho center of symmetry of a flat, homogeneous object is the center of mass, find (he center of mass of a square; of a rectangle; regular hexagon; circle. 15. It is evident that if the medium of a triangle (BM) be placed on a knife edge the triangle will balance (for if P!" be II AC, the pull on P is balanced by the pull on P'). Hence find the Renter of mass for any triangle. Tor a regular pentagon. It is tiscfui to be able Id determine i.he center of mas-s of an object by neomi^try or by any other meaus, sincii a knowledge of the center of mas.s ot a Ixidy often pi\!il)lis us to treat the body in a simple way, tor example, as il llie l.o(l> iviTc concentrated at a single point. y Google 5^0 nEOMKTHY 16. if a box has a triaaguiar end, subject to thi points, at. u'liat single poict oa the end must a be applied? 17. The cross section of a cylinder is a circle. The woiglit-supporting strength of a horizontaJ cylindrical beam of given material and length varies as the area of the cross section times its radius. Compare the weight-supporting power of two solid horizontal iron cylindrical beams of the same length and quality of iron, the radii being 3 in. and 6 in. respectively. (PoinI out and use the short way of getting the desired result.) IS. The cross section of a hollow cylinder (i.e. of a tube) is a circular ring. Denote the outside radius of the tube by R and the inside radius by r. Then it may be shown that the weight-supporting power of a hollow cylindrical tube of given length and material varies as the area . , . iP + r= of the cross section (ve. of thermg) times — -- — . It R = i in, and r = 3 in., compare the weighl-wiipporiinir powf^r of the tube, with that of a solid cylindrical beam of the same length and same cross sectional area. 19. Make up and work a similar example. In general a cylindrical tube is stronger than a solid cylindrical beam of the same length and containing the same amount of material. Hence in a framework, as in that of an jurship where the maximum strength must be obttuned from a given amount of material, the metallic rods and posts iire all tubular. In like manner, bamboo rods, since theyai'e hollow, are used in an aeroplane instead of solid wooden rods wherever possible. For the same reasoB, the bones of flying birds, and many bones in men and animals are hollow and not solid n 20 To coni^trucf a square which '.hail be ap- proMm'itely equivalent to a given circle divide the radiun OA into four equj,! parts and produce each end of two perpendicular diameters a ell's ; equal to one fouilh of the ra liiis and rjnnect the extremities of the lines thus formed slio« that taking the squaie thus foimed aa equn ilent t the circle i= th( sime as taking " = 3J, Also find th( per <u!t of erroi in tikmg thi quare aa equivalent to the circle 21. A short wiy to construct a regular inscribed pentjg n ^.n I aho il five-pointed stii l w | Liitagrarr ) i-. aa followa Dii \ a ii le and y Google PUACTICAL APPLICATIONS 531 two diamelors /IB atiti CD at right angles. Bisect the radius OB at F, and with f as a center and FC as a radius describe an arc cutting AO at H. Then CIl is the length of a side of the regular inscribed pentagon, by joining whose alternate vertices the pentagram niay be formed. As a proof of this, by use ot right tri- angles, prove pVio- 2V-5 CH - fl— C^ee E\ 1" p 301 ) 22 f irp iitpra some lies use the follow mg ^ thod of approximately deteimining the length t a circumference whose ladiu'i is Inown: Let O be the giien circle Dran Oi ind OB, radii t right angles Draw AB j,n1 the radius OELAB ,t D Measure DE Then take cncuraference ^ =b iO + DE Find the per cent of erroi in this method, taking t = 3 14159-. 23. The following designs are important in architecture or in orna- mental work (thus the first is a detail in a stained glass window in an early French cathedral; the second is the plan of the base of a coS- umn in an English cathedral). Discover a method of constructing each of tiic following designs, and reproduce each of them in a drawing: y Google 532 CEOMETRY 21. Construct a square. Take each vertex of the square as a ccater and one half a side of the square as a radius and describe arcs which meet. Erase the squai'e and you have a quatrefoi!. By drawing other circles ajid arcs of circles, elaborate the quatrefoil into an orna- mental design (see design 11, p. 520). 26. In like manner consli'uct a cinquetoil by use of a regular penta- gon and develop it into an ornamental design. 26 Troj,t I icgular JicMgon m the same way. 27 4 pa^tmeiif or raouiL may be formed out of regular polygons in thf, folloivmg '^^^b Shou how to m:i.kc each diagram in the simplest y Google APPLICATIONS OF SOLID GEOMETRY TO MECHANICS AND ENGINEERING EXERCISES. GROUP 93 (Books \'I and VII) a the fl:it 1 A carpenti edge to the surface in various diriK;t; the flitne^s of a will surface? Wha these mechanics' 2 EJipHin nhj in object with I of a surface by applying a ; (might w does a plasterer test. ic principle is used by I'eii legs, as a stool or tripod, object with tour legs, iis a table, ever use foiar-leggcd pieces .of furniture? 3. How can a earpt;nter get a comer post of a house in a vertical position by use of a carpenter's Btiuaro? What geometrical prineiple does he 4. The diagram is the plan ot a hip roof. The slope of each face of the roof is 30°. Find ihe ienglh of a hip rafter sAB. 1 (he pli trianRle DBC repre,wnli Hence it may l>e fih El section ot the roof at 20 I ISC = - v;!. In like manner by taking a section through BF, it is found that AC = 10. Hence in %e triangle ABC, AB may be found.] 6. Find the area of the entire roof represented in Ex, 4, 6. Make drawing showing at- wiiat fttigle Ihe two ends of a rafter like liC in Ex. 4 y Google 5^4 GEOMETRY 7. Make drawings ?liowinR at what angle a jack raftpr like 12 in Ex. 4 rauBt be cut. ISuiJ. To determine how the end 1 of the jaek rafter must be eut use the principle that two intersecting straight fines determine a plane (Art. 501). The cutting plane at 1 must make an angle ;it ihe side of the jack rafter equal to angle CBH, and on the top of the jiick rafter equal to angle ABC] 8. What is a gambrtjl root? Make up a set of examples coticorning a gambrel roof similar to Exs. 4-7. 9. By use of Art. 615, show that a page of this book held at twice the distance of another page from the same lamp rei'ei\'es one fourth the light the first page receives. 10. The supporting power of a wooden beam varies directly as the area of the erosa Heotion times the height of the beam and inversely as the length of the beam. Compare the supporting power of a beam X2 ft. long, 3 in. wide, and 6 in. high with that of a. beam 18 ft. long, 4 in, wide, and 10 in. high. -Uso compare the volumes of the two beams. EXERCISES. GROUP 94 (Books VIII and IX) 1. A hollow cylinder whose inside diameter is 6 in, ia partly filled with water. An irregularly shaped piece of ore when placed in the water causes the top surface of the water to rise 3.4 in, in the cylinder. Find the volume of the ore. 2. What is a tubular boiler? What is the advantage in using a tubular boiler as compared with a plain cylindrical boiler? If a tubular boiler is IS ft. long and contains 32 tubes each 3 in. in diameter, how much more heating surface has it than a plain cylindrical boiler of the same length and 36 in. in diameter? (Indicate both the long method and the short method of making this computation and use the short method.) 3. If a bridge is to have its linear dimensions lOOO times as great as those of a given model, the bridge will be how many times as heavy as the model? Why, then, may a bridge be planned so that in the model it will sup- port relatively heavy weights, yet when constructed according to the model, falls to pieces of its own weight? Show that this principle applies to other ronstructions, such as buildings, machines, etc., as well as to bridges. y Google rRACTICAL APPLICATIONS 535 i. Work again Eks, 23-25, 27, p. 473. 6. Make and work for yourself an example similar to Ek. 24, p. 473. 6. SouniJ spreads from a center in the form of tlie surface of an expanding s[)here. At the distance of 10 yd. from the source, how will the surface of this sphere compare with its surface as it was at 1 yd.? How, then, does the intensity of sound at 10 yd. from the source com- pare with its intensity at a distance of 1 yd.'f Does this law apply to all forces which radiate or act from a center aa to light, heat, magnetism, and gravitation? Why is it called the law of inverse squai'es? 7, If a body be placed within a spherical shell, the attractive forces exerted upon the body by different parts of the shell will balance or caned each other. Hence a body inside the earth, as at, the foot of a mine, ia attracted effectively only by the sphere of matter whose radius is the distance from the center of the earth to the given body. Hence, prove that the weight of a body below the surface of the earth varies as the distance of the body from the center of the earth. [Stjg. If W denote the weight of the body at the surface, and w its weight when below, R the radius of the earth, and r the distance of the body from the center when below the surface, show that 8 The heht ol tl e sun falling on a smaller sphere, as on the earth oi the moon cau-,es that b^dv to cast a conical shadow. Denoting the radium of the sun bj R the radius of the smaller sphere by r, the diatanct between the two spheres dr by d, and the length of the shadow by I, show that I = - _ ■. Find I when d - 92,800,000 mi,, R = 433,000 mi., r = 4000 mi. 9. If in a lunar ecli|}se the moon's center should pass through the axis of the conical shadow, and the moon is traveling at the rate of 2100 mi. an hour, how long would the total eclipse of the moon last? How long if the moon's center pas.?ed through the earth's conical ahadon- at n distance of 1000 mi. from the axia of tlie cone? y Google 531; c i)':oMr,'ri!v 10, If the moon's diaraotcr is 2160 mi., finil Ihn length o( the moon's shadow us caused by sunliglit. 11. If the distance of the moon from the earth's eenter VMrics from 221,600 mi. to 232,970, show bow this explains why some ei;lipscs of the mm are total and others annular. Why, also, at a given point on the earth's surface is an eclipse of the sun a so much rarer sight than an eclipse of the moon? Why, also, is its duration so much briefer? y Google FOEMi.n.AM OF" PLAICE OEOMETEY SYMBOLS 6, <!=siiles of Uianglo J/tr. /,'--^iadiH3 of a tilrcle. s = i{a + b + c). i).-diameter of acirele. ftf = aUitude on sklo r. (' = cirpiimferenee of acirele. 5ii( = meiliiin on side ('. f = i'iitlius of an inso'lbed *c=biaettorotanglB.>i.!"Hile pirole. SLiii.' .-. T = -r-apprax. (or :i. 141(3— ). id)n=liiiB segnjouLs. K=area. /■= [loi'iuietcr. /i = liase of a triangle. 6\i = slde of a regular polyi^oii h =altitude of a triangle, ol » sides. ''[ anil ?i- = bases of a trapezoid. LENGTHS OF LINES 1. In a right triangle, C being the tight angle, f' = n' + &'. Art. 346. '2. In a riglit, triangle, ( and m being Ihe projections of a and fe f, and li, tho altitude on c, 3. In an oblique triangle, m being tho projectioo of b on e, if aisopposite an ai^^ito Z, a' = b= + c= — 2 cX m. Art. 319. if a is opposite an obtuse Z, a' = 6' + e' + 2 cX*"- Art. 350. ».-iJ/<«-<.)(--»)l— «J Art. 393. me=iT/at«^ + 6"j-c^ Art. 353. If I and m are the segments of opposite, a:J' = l : Art. 3li3„ c made by the biaeotor o( the m. Arts, 333, 336. y Google ISDEX OF DEFiXlTIOXS ASD FOEJIULAS ^bbrp It on ilgebr ic analy 3 Algebra e u eti o ■ilternat on Altitude of eone of frustum f of frustum of p of parallelogr n otpr.m of pyrom d of spher c I of trapt of tr ngle of zone Anil J olu o An{,!c at CD to diled formed hv a etra ght I n formed bv t inscr bed n I Angle of lull poljhedrnl right salient , sides of . splierieal straight tetraliedial trihedinl vertex ot Angles, adjacL'i alteniato-iiili llOH)ologOll;^ . interior , . of polygon of quadriliiti of spherical Appolom yGoosle IND]:X OF IH'.FINITIOXS AXD FOTiMlXAS major , . Archimedes Arcs, eonjligiite Area ol surface Aryabhatta August, E. F. Auxiliary linea Axioms, generiil , geometric Axis of circle of splie of circular cone of regular pyiaoiii; of symmetry Babylonians . Base of cone . of isosceles triai of pyramid of spherical pyr; of spherical sect of triangle , ! of < lulcr of fnislum of or of frustum of ]ii of parallelogram of prism of spherical segment . of trapezoid of zone . . Bisector of triangle . Bodies, the Three Round Center of circle . of regular polygon of sphere .... -'■' Chord Circie 10 are of center of ... . circumference ul". tormiil: for circumscribed . . . formula for area of 103 ,406 103 103 103 274 103 103 inscribed .... radius of ... , sector of ... , small Circles, conecnliic . . escribed .... 105 10.3 104 427 ID.i rireum-centcr of Irmn^'le Civcumtercncp . . . IJ Classification of ]>islvln-dion of triangUM . . . Complement .... Composition . , , . 81 .103 3G0 2,33 125 16 181 Conclusion .... Coninirrent lines . 24 80 Cone altitude of ... ^^xis of base Of circular circular, fotiiiuhi for vo uine of ... . circular, formulas for la eral and total area . 412 412 413 413 418 417 y Google INDEX OF DEFINITIONS AXD FORJIULAS Cone, elements of lateral surface of . oblique circular of revolution . right circular . vertex of Cones, similar Congruent figures Conical surface . directrix of element of . generatrix of . vertex of . . Conjugate area Conaequenta Constant . geometrical . Constants . Continued proportioi Continuity, principle of . Converse of a th< Corollary . Cube . . . Curved spaces Curved surface Cylinder , altitude of . bases of circular . circular propei elements of . lateral surface of . oblique of revolution of revolution, formulas fc lateral and total arc; of . . . right . . . right ciroula. Cylinder, right Cylinders of re' Cylindrical sut elor t of . D t mn d Pl D g n 1 pol K f p lyl J f q d 1 t D m t 1 t ph Dh d 1 1 y Google I^■DEX OF DEFIXmOXS -VXD FOIi:\IUIA Duality, Fi^ ■ipk of Edge o£ dihedral ni, Edges of polyhedral of polyhedron . Egj-ptiiins . , . 400, 401, 432, Elements of conical f. of cylindrical surface . Enunciation, general pftiticular Equal figures . Equivalent figures . solids .... Euclid Eudosua . . .493 Euler European .... Extreme and mean ral Extremes .... Faces of dihedral angle , of poiyhedTiil angle of polylirdmn . Figure, curviliiiuar geometrical . mixtilinear . plane_ . . . Figures, roctilinoar Formulas of Piano Geo net for lengths of 1 ne j of Plane Oeoniet fo areas of plane fi re^ of Solid Geometry fc of Solid Geometry ft volumes , Foot o£ perpendicular Fourth propoilionul PruBtuni of toiiu. altitude of . , bases of . . . formula for l.ileral of .... formula for volume i lateral surfacp of , slant height of Frustum of pvraniid altitude of . . . base of slant height of ;al ! of c.rlintirical sui Geometric solid , Geometry . epochs in development of 4 history of . modern . Non -Euclidean , Jrigu of e ka 400 102, 403, 494 Ha n on i . , . 108 Hept oon .... 74 He o . . .496 He 1 on ... 74 He h i on . . .360 H n loos 400, 432, 497 Hpp U3 , . .408 H ppoc atus 491, 492, 406 Hstor ofg t 490-498 Honol go 3 a =1 .... 34 y Google IXDEX OF DEFINITIONS AXD FOKJIULAS Homology, Principle ot 487 I ne d ided tein^llv Hyperapafie .... 48S 1 ot centres Hypoteniiae .... 33 I rallel to pi Hypothesis .... . 24 I rj end eul r t j 1 Icosahedron .... , 3G0 ° In-center of triiinplE! . s; L a au 1 irv Inclination of line to pia e 347 concurre t . 1-25 Lobatche 1 ratio . 1:25 Loci^ Inference, immediate . . 23 Log eil U t)l Inversion . 180 Lune Isoperi metric figures . 280 inn-lp ot for i f Jews , 43- apler al degr es torn nU for or a Lateral nrea of frustum ol square un ts ot a pyramid, formula for . 379 ot pyramid .... . 37 Alagn tude n Lateral edges of prism 3G1 mcommen 1 1 of pyramid 377 Alaji nun Lateral faees of prism ill "Mean p opo t a 1 of pjramid 3" Mea Lateral surfaoe uf cnne 41' Me h n cal n et} 1 o£ ej Under 401 M d an of trapezo 1 of frustum ot eon-^ 414 ot tranje Legs of iso.^d<^luaa> 3 "\Ienela s of riplit trnn„-lL 31 Method ilgebi of trapwoid Iinut (1 \1 of 1 ts Method loj, c 1 Limits, metiiol of \1 meeh i -x] Line 1 ot n I broken 11 tion« curbed n hcto e 1 divided exttnmllj 1')^ Metr e sjste dnided hatmonii ,\h I'lS ^\ nor n <■ ot <■ 1 di\ided in p\titiiic i ii ■\I te y Google INDEX OF DKriMTiONS AN)") FOILMULAS Nappes of cono . Negative quaniili™ Non-Eup]idean gcoi Numerical t Octagon Octahedron Opposite of a thci Origin of geoiiK'tr Ortlio-eenter Parallei lines Parallelogram altitude of . bases of Parallelopiped right . . rectangular Parallel plants Pentagon . Pentedecagon . Perimeter of p'll of triangle . Perpendicular liii planes Pi {«) . . Plane . . . determination Pianes, parallel Plato . . , Point . . . PointB, coney lie Polar distance of triangle , , Poljhediai angles, equal synimetTical Polyhedron diagonal of edges of , y Google IKDEX OF DEFINITIONS AND FOT!ML"LAS Polyhedron 1 1 Polyhedrons classificiti n Postulnfi. of solid gpon Postulites logical Prism altituclc if base of oirpumsLribed aboi inscribed in oil! lateral area o! lateral edges of lateral fates of oblique quailrdnjrular right Tight sectirn i trianguhr tmncuteJ Pri'imaioid bases of . formula for voli Prismoid . Problem Pi ejection of lint of line on plan of point on line of point on plar Proof by auperpciiiio i forms of method? 1)1 proportion eontinu.d , . torins of . . third axi ,f base of circumscriljei about fiustum of jnbcniel in on latcnl ed„ ol lalor fie qmdr nfnil r regular regular slint 1 tif,! spherical trnngul r truncated verte\ of PTtha' 491, 492, 4!)S, 4!i7 Q d t f Qu d 1 t 1 angl f n ! tud II3 p i>ortionaI 1 p neipic of y Google IXDEX OF DEFINIT10^:S AND FORMLXAS Rhombus (it) f.o!id^ phTBin! Right section of <L hnloT 4f)5 Spaei-- cuned of pusm 3bl Sphere romaiis 4 4 4')0 ftilS of Round liiijic Tl t n rep a-^s center of Eeholium 23 Lircuniscrihtd \bna Vp 1 t line 104 hedron Bfond 17 diameter of Sect 14 formula fot arpx Election o! I 1 ilJ un J1)0 face of Sector of cink 104 formula for ^ohlnl of cirplo fori lU fjr great circle of of spilt re 4bl inscribed in polili Sectors amiJ.r 275 poles of Segrapnt of i tlo 114 radma of of line 14 small circle of of sphere 4b2 Spherical angle Segments un ilar 2T3 degrees Semicircle 101 Seniicireumfeceii"e 103 SpheriPil poh„ua ''ides of angle 14 angles of of poljgon 73 sides of of quadiilateril u6 vertices of of triangk 32 Sphericil piramid Similar cones of r oluti on 113 base of eUmdeis oi i ol li ii 404 vertex nf polvgina (i3 130 Spherical sector base of pohhedrons J03 formula for lolum Btttors of circlps Spherical segment segments of civ(1p9 2-j altitude of Slant height of coik 413 of frustum of cone 414 formulis tor loluu of frustum of pyramid of one bnie of regular pyr-iinid ^77 Spherical triangle Solid bi- rectangular geometry .... 18,319 formula for area Solids, equivalent . . . 303 sqnnre unifs of geometric .... . 11 tri-rpcfangular y Google INDEX OP DEFINITIONS AND FOEilULA; Spheiicil trim, mtntal Bymmetrici 1 Spheneal wedgt. Spherid Stiaight line Superposition Supplement area of conical Tangent circles langent conimon evternil Ime to tpleie plane to cjlinJer plane to &phei Terms of a proportioi Tetrahedral angle Tetrahedron . Thales . , Theorem Theory of liniil Third proport: Transversal Trapezium Trapezoid . altitude of . Iriangl d' 7-1 acute 33 altituilts jt 33 base of 33 bisectors of 34 bisectois of forniU for 213 clagaifications ut 3', 33 equiangular 33 equilateral S2 isosceles 32 medians of 34 oltuse 33 pohr 441 ii^ht 33 icilcne 32 spheueal 438 \crfc\ angle of 33 \eitex of 33 riihedral angle 349 bi rectangulir 330 isosceles 350 lectanguUr 350 tiirtotangular 350 Ungula .... . dlil Unit of measure . - of surface . . . . 231 of volume . . . , . 3S3 Units of angle . . , 17 of spherical surface . 452 Variable .... . . 120 Vertex ani^le . . . 33 Vertex of angle . . . . 14 of polyhedral angle . , 349 of pyramid . , . . 377 of spherical pyramid , 461 y Google IXDliX OF nEFIXITlOXS AND FORMULAS Vut'teic of triangle . Vertices of polygon . of polyhedron . of quadrilateral of spherical polygo of triangle . Volume of solid . y Google